Health Education Systems Inc (HESI) A2 Exam Quiz: Gravity And Free Fall
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Gravity And Free FallQuestion 1 of 14

A canister is launched vertically upwards from the ground with an initial speed of 30 m/s. Neglecting air resistance, what will be its speed when it returns to its starting point on the ground?

0 m/s
15 m/s
30 m/s
60 m/s
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Health Education Systems Inc (HESI) A2 Exam Quiz

Health Education Systems Inc (HESI) A2 Exam Quiz: Gravity And Free Fall

Practice Gravity And Free Fall in Health Education Systems Inc (HESI) A2 Exam with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Gravity And Free Fall, giving you a quick way to practice the rules, question types, and explanations that matter most for Health Education Systems Inc (HESI) A2 Exam.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A canister is launched vertically upwards from the ground with an initial speed of 30 m/s. Neglecting air resistance, what will be its speed when it returns to its starting point on the ground?

  1. 0 m/s
  2. 15 m/s
  3. 30 m/s (correct answer)
  4. 60 m/s
Explanation: This question tests your understanding of projectile motion and energy conservation principles that appear on physics-related sections of standardized exams. When analyzing vertical projectile motion, you need to recognize that gravitational acceleration works symmetrically. As the canister travels upward, gravity continuously slows it down at 9.8 m/s29.8 \text{ m/s}^2 until it reaches maximum height where velocity equals zero. During the downward journey, gravity accelerates the canister at the same rate of 9.8 m/s29.8 \text{ m/s}^2. Since the canister returns to the same height (ground level) where it started, it will have the same speed it began with—just in the opposite direction. You can also think of this through energy conservation: the initial kinetic energy converts to gravitational potential energy at maximum height, then converts back to the same amount of kinetic energy when returning to the starting position. Answer choice A (0 m/s) represents the velocity only at the maximum height, not when returning to ground level. Answer choice B (15 m/s) might tempt you if you incorrectly think the speed is somehow halved during the round trip. Answer choice D (60 m/s) incorrectly suggests the speed doubles, perhaps from misunderstanding how gravity affects motion. The correct answer is C (30 m/s)—the canister returns with exactly the same speed it started with. Study tip: For any projectile motion problem involving a return to the starting height, remember that speed at launch equals speed at return. This symmetry principle will save you time on calculations.

Question 2

A nurse accidentally drops a 10 g vial and a 100 g syringe from the same height at the same time in a patient's room. Ignoring the effects of air resistance, which statement accurately describes their motion?

  1. The 100 g syringe will hit the ground first because its greater mass results in a stronger gravitational pull.
  2. The 100 g syringe experiences a greater gravitational force, but both items hit the ground simultaneously. (correct answer)
  3. Both items will accelerate at different rates that are directly proportional to their respective masses.
  4. The 10 g vial will hit the ground first because it has less inertia and is therefore easier to accelerate.
Explanation: When you encounter physics problems about falling objects, remember that gravity affects all objects equally regardless of their mass - this is one of the fundamental principles discovered by Galileo. Both the vial and syringe fall with the same acceleration (g=9.8 m/s2g = 9.8 \text{ m/s}^2) and hit the ground simultaneously. While the heavier syringe does experience a greater gravitational force (F=mgF = mg), it also has proportionally greater mass to accelerate. These effects cancel out perfectly: a=F/m=mg/m=ga = F/m = mg/m = g. The acceleration is identical for both objects. Choice A reflects the common misconception that heavier objects fall faster. While the syringe does experience stronger gravitational pull, its greater mass requires proportionally more force to accelerate, so the net effect is zero difference in acceleration. Choice C incorrectly suggests that acceleration varies with mass. In reality, gravitational acceleration is constant for all objects in the absence of air resistance - this is why a feather and hammer dropped on the moon (no air resistance) hit the surface simultaneously. Choice D misapplies the concept of inertia. While the vial has less inertia (resistance to acceleration), it also experiences less gravitational force. Again, these factors balance out perfectly. For HESI physics questions, remember that gravitational acceleration is always the same for all objects when air resistance is ignored. This counterintuitive concept frequently appears on standardized exams because it challenges our everyday observations where air resistance does affect falling objects differently.

Question 3

A crumpled piece of paper and a flat sheet of paper, both with identical mass, are dropped from a second-story window. Which of the following best explains the observation that the crumpled paper hits the ground first?

  1. The crumpled paper has a greater effective density, which causes it to overcome gravity more efficiently.
  2. The flat sheet experiences a greater upward force from air resistance due to its larger surface area. (correct answer)
  3. The force of gravity on the crumpled paper is significantly greater than the force on the flat sheet of paper.
  4. The crumpling process reduces the paper's inertia, allowing it to accelerate more quickly through the air.
Explanation: When objects fall through air, you need to consider both gravitational force and air resistance. Both forces determine how quickly an object reaches the ground. The crumpled paper hits the ground first because air resistance affects the two papers differently, even though they have identical mass. The flat sheet has a much larger surface area exposed to the air as it falls, creating significantly more air resistance (drag force) that opposes its downward motion. This upward force from air resistance slows the flat paper's fall. The crumpled paper, with its compact shape and smaller surface area, experiences much less air resistance and falls closer to its natural acceleration due to gravity. Choice A incorrectly suggests density affects how objects "overcome gravity." While the crumpled paper is denser, density doesn't change how gravity acts on an object - gravitational force depends on mass, which is identical for both papers. Choice C is wrong because gravitational force depends only on mass (F=mgF = mg). Since both papers have the same mass, gravity pulls on them with exactly the same force. Choice D misunderstands inertia, which is an object's resistance to changes in motion and depends only on mass. Crumpling doesn't change the paper's mass, so its inertia remains the same. Remember: When analyzing falling objects in real-world conditions, always consider air resistance. The shape and surface area of objects significantly affect how air resistance opposes their motion, while gravitational force depends solely on mass.

Question 4

According to Newton's Law of Universal Gravitation, if the distance between two objects is tripled, while their masses remain unchanged, the gravitational force between them will:

  1. decrease to one-ninth of the original force. (correct answer)
  2. decrease to one-third of the original force.
  3. increase by a factor of three.
  4. increase by a factor of nine.
Explanation: Newton's Law of Universal Gravitation states that gravitational force is proportional to the product of two masses and inversely proportional to the square of the distance between them: F=Gm1m2r2F = G\frac{m_1m_2}{r^2}. The key insight here is that distance appears in the denominator as a squared term. When you triple the distance (multiply by 3), you're substituting 3r3r for rr in the equation. This gives you Fnew=Gm1m2(3r)2=Gm1m29r2F_{new} = G\frac{m_1m_2}{(3r)^2} = G\frac{m_1m_2}{9r^2}. Since the original force was Foriginal=Gm1m2r2F_{original} = G\frac{m_1m_2}{r^2}, the new force becomes 19\frac{1}{9} of the original force. Choice A correctly identifies this one-ninth reduction. Choice B incorrectly assumes a linear relationship where tripling distance simply divides force by 3, ignoring the inverse square relationship. Choice C suggests force increases by a factor of three, which completely misses that gravitational force decreases as distance increases. Choice D suggests the force increases by a factor of nine, which not only gets the direction wrong but also misapplies the squared relationship. Remember the inverse square law pattern: when distance changes by a factor of nn, force changes by a factor of 1n2\frac{1}{n^2}. This relationship appears throughout physics (electromagnetic forces, light intensity, sound intensity), so mastering it will help you with multiple question types on science exams.

Question 5

Two identical stones are released from a tall building. Stone A is dropped from rest. At the same instant, Stone B is thrown vertically downward with an initial speed of 10 m/s. How do their accelerations compare just after release, assuming no air resistance?

  1. The acceleration of stone B is greater than that of stone A.
  2. The acceleration of stone A is greater than that of stone B.
  3. Both stones have the same non-zero acceleration. (correct answer)
  4. Both stones have zero acceleration at the instant of release.
Explanation: When you encounter problems involving objects falling under gravity, the key principle to remember is that gravitational acceleration affects all objects equally, regardless of their initial motion or mass. Both stones experience the same gravitational force per unit mass, which means they both have the same acceleration of g=9.8 m/s2g = 9.8 \text{ m/s}^2 downward immediately after release. The initial velocity of an object doesn't change the acceleration it experiences due to gravity. Stone A starts from rest (0 m/s) and Stone B starts with 10 m/s downward, but gravity pulls on both with identical acceleration. Let's examine why the other options are incorrect. Option A suggests stone B has greater acceleration because it was thrown downward, but this confuses initial velocity with acceleration—the throwing motion only affects starting speed, not the gravitational pull. Option B incorrectly implies that starting from rest somehow creates stronger gravitational acceleration, which isn't physically possible. Option D states both stones have zero acceleration, which would only be true in a gravity-free environment or if they were floating in space. The correct answer is C: both stones experience the same non-zero acceleration due to gravity, regardless of their different initial velocities. For physics problems on standardized exams, remember that gravity is the great equalizer—it accelerates all objects at the same rate near Earth's surface. Don't let initial conditions like starting velocity trick you into thinking acceleration changes. Focus on what forces are acting on the objects.

Question 6

A student weighing 500 N stands on the ground. According to Newton's Third Law of Motion, what is the reaction force to the Earth's 500 N gravitational pull on the student?

  1. The 500 N upward normal force exerted by the ground on the student.
  2. The 500 N gravitational force exerted by the student on the Earth. (correct answer)
  3. The frictional force between the student's shoes and the ground, which is also 500 N.
  4. There is no reaction force because the Earth is too massive to be affected by the student.
Explanation: When you encounter Newton's Third Law questions, focus on identifying true action-reaction pairs - forces that are equal in magnitude, opposite in direction, and act on different objects. Newton's Third Law states that for every action, there is an equal and opposite reaction. The key is understanding what constitutes a proper action-reaction pair. Here, the Earth exerts a 500 N gravitational force downward on the student (the action). The reaction must be the student exerting an equal 500 N gravitational force upward on the Earth. This is exactly what choice B describes - the gravitational force the student exerts on the Earth. Choice A incorrectly identifies the normal force from the ground. While this force does equal 500 N upward (balancing the student's weight), it's not the reaction to Earth's gravitational pull. The normal force is a contact force from the ground, not a gravitational force, and it acts on the student, not on the Earth. Choice C mentions friction, which is irrelevant here. Friction acts horizontally and wouldn't equal the student's weight unless the student were somehow sliding at an angle. This completely misses the gravitational action-reaction pair. Choice D reflects a common misconception that massive objects don't experience reaction forces. While Earth's acceleration due to the student's pull is negligible, the force itself still exists and equals 500 N according to Newton's Third Law. Remember: Action-reaction pairs always involve the same type of force acting on different objects. Don't confuse them with balanced forces acting on the same object.

Question 7

A medical device is thrown vertically upward with an initial speed of 20 m/s. How much time does it take to return to the hand of the healthcare worker who threw it? (Use g = 10 m/s² and neglect air resistance).

  1. 2 seconds
  2. 4 seconds (correct answer)
  3. 6 seconds
  4. 8 seconds
Explanation: When you encounter projectile motion problems in healthcare contexts, remember that objects thrown vertically follow predictable patterns governed by gravity, regardless of whether it's a medical device or any other object. For vertical motion, you can use the kinematic equation: v=v0−gtv = v_0 - gt, where the final velocity vv equals zero at the highest point. With an initial velocity of 20 m/s upward and gravity of 10 m/s² downward, the time to reach maximum height is: 0=20−10t0 = 20 - 10t, so t=2t = 2 seconds. Here's the key insight: the total flight time is twice the time to reach maximum height due to symmetry. The device takes 2 seconds to go up and another 2 seconds to come back down, giving a total time of 4 seconds. This makes B) 4 seconds correct. Looking at the wrong answers: A) 2 seconds represents only the time to reach maximum height, not the complete round trip. This is a common trap where students calculate halfway through the problem. C) 6 seconds and D) 8 seconds are too large and likely result from calculation errors or misapplying formulas. You could also solve this using h=v0t−12gt2h = v_0t - \frac{1}{2}gt^2. Setting height h=0h = 0 (returns to starting point): 0=20t−5t20 = 20t - 5t^2, which factors to t(20−5t)=0t(20 - 5t) = 0. This gives t=0t = 0 (start) and t=4t = 4 seconds (return). Remember: for vertical projectile motion, always consider whether the question asks for time to maximum height or total flight time—they're very different values.

Question 8

A medical package is dropped from a helicopter that is moving horizontally at a constant velocity. From the perspective of a stationary observer on the ground, what is the horizontal motion of the package as it falls (ignoring air resistance)?

  1. The package immediately slows down horizontally due to gravity.
  2. The package maintains the same horizontal velocity as the helicopter. (correct answer)
  3. The package accelerates horizontally in the helicopter's direction.
  4. The package drops straight down with zero horizontal velocity.
Explanation: When you encounter physics problems involving projectile motion, remember that horizontal and vertical motions are independent of each other. This is a fundamental principle that applies when objects are dropped or thrown from moving platforms. The key insight here is Newton's first law of inertia: an object in motion stays in motion unless acted upon by an external force. When the package is dropped from the helicopter, it already possesses the same horizontal velocity as the helicopter. Since we're ignoring air resistance, no horizontal forces act on the package after it's released. Therefore, the package maintains its horizontal velocity throughout the fall while gravity only affects its vertical motion. Let's examine why the other options are incorrect: Option A suggests gravity affects horizontal motion, but gravity only acts vertically downward. It has no horizontal component to slow the package's horizontal movement. Option C implies some force accelerates the package horizontally, but with air resistance ignored, no such force exists once the package is released. Option D represents a common misconception that dropping something means it loses all horizontal motion. This would only be true if the helicopter were stationary. The package follows a parabolic trajectory: constant horizontal velocity combined with increasing downward velocity due to gravity. To an observer on the ground, the package appears to fall in an arc while maintaining the helicopter's forward speed. Study tip: For HESI physics questions involving projectile motion, always separate horizontal and vertical components. Remember that gravity only affects vertical motion, while horizontal motion continues unchanged unless other forces intervene.

Question 9

A 2 kg object is dropped from a height of 10 meters. Just before it hits the ground, its kinetic energy is approximately 196 Joules. If the same object were dropped from 20 meters, what would be its approximate kinetic energy just before impact, ignoring air resistance?

  1. 196 Joules
  2. 294 Joules
  3. 392 Joules (correct answer)
  4. 784 Joules
Explanation: When you encounter physics problems involving falling objects, you're dealing with energy conservation principles. The key insight is that gravitational potential energy converts to kinetic energy as an object falls. The relationship between height and kinetic energy is direct and proportional. Since kinetic energy equals the potential energy lost during the fall, we can use the formula KE=mghKE = mgh, where m is mass, g is gravitational acceleration (9.8 m/s²), and h is height. From the first scenario, we can verify: KE=2 kg×9.8 m/s2×10 m=196 JKE = 2 \text{ kg} \times 9.8 \text{ m/s}^2 \times 10 \text{ m} = 196 \text{ J}, which matches the given value. For the 20-meter drop, the kinetic energy becomes: KE=2 kg×9.8 m/s2×20 m=392 JKE = 2 \text{ kg} \times 9.8 \text{ m/s}^2 \times 20 \text{ m} = 392 \text{ J} This confirms answer C is correct. Looking at the wrong answers: A (196 Joules) incorrectly assumes height doesn't affect kinetic energy. B (294 Joules) represents a common error of adding 98 J (half the original energy) rather than doubling the total. D (784 Joules) mistakenly squares the height relationship, treating this like an area calculation rather than a linear energy relationship. Study tip: Remember that gravitational potential energy (and thus kinetic energy at impact) scales linearly with height, not exponentially. When height doubles, kinetic energy doubles. This direct proportionality appears frequently in HESI physics problems involving energy conservation.

Question 10

A person stands on a scale in an elevator. The scale reads their normal weight of 700 N when the elevator is stationary. If the elevator begins to accelerate downwards at 2.0 m/s22.0 \ m/s^2, what will the scale read? (Use g=9.8 m/s2g = 9.8 \ m/s^2)

  1. 557 N (correct answer)
  2. 700 N
  3. 843 N
  4. 0 N
Explanation: When you encounter elevator physics problems, you're dealing with apparent weight versus actual weight. The key insight is that a scale measures the normal force pushing up on you, not your true weight. When the elevator accelerates downward, you need to apply Newton's second law. Your actual weight (gravitational force) remains 700 N downward, but now you're also accelerating downward at 2.0 m/s². First, find your mass: m=700 N9.8 m/s2=71.4 kgm = \frac{700 \text{ N}}{9.8 \text{ m/s}^2} = 71.4 \text{ kg} Now apply Newton's second law in the downward direction (taking downward as positive): mg−N=mamg - N = ma 700−N=(71.4)(2.0)700 - N = (71.4)(2.0) 700−N=142.8700 - N = 142.8 N=557 NN = 557 \text{ N} The scale reads 557 N, confirming answer A is correct. Looking at the wrong answers: B (700 N) assumes the elevator's acceleration doesn't affect the scale reading, ignoring that apparent weight changes with acceleration. C (843 N) incorrectly adds the acceleration effect instead of subtracting it—this would be the reading if the elevator accelerated upward. D (0 N) would only occur in free fall, where acceleration equals g, not the given 2.0 m/s². Remember this pattern: downward acceleration decreases apparent weight, upward acceleration increases it. The scale always reads less than your true weight when accelerating downward because you're "falling" along with the elevator, reducing the force you exert on the scale.

Question 11

A skydiver jumps from a plane and falls for several seconds before opening her parachute. She reaches a constant speed, known as terminal velocity. At this point, which statement is true?

  1. The force of gravity acting on her is effectively zero at that altitude.
  2. Her acceleration is constant and equal to g (approximately 9.8 m/s29.8 \ m/s^2).
  3. The upward force of air resistance is equal in magnitude to the downward force of gravity. (correct answer)
  4. Her inertia becomes zero, preventing any further change in her state of motion.
Explanation: When you encounter physics questions about objects moving at constant velocity, focus on Newton's First Law and the concept of equilibrium. Terminal velocity occurs when all forces acting on an object are balanced, resulting in zero net force and zero acceleration. At terminal velocity, the skydiver has reached a state where the upward force of air resistance exactly equals the downward gravitational force. These two forces cancel each other out, creating equilibrium. With no net force acting on her, she stops accelerating and maintains a constant downward speed. This is why option C is correct - the forces are equal in magnitude but opposite in direction. Let's examine why the other choices are incorrect. Option A suggests gravity becomes zero at altitude, but gravitational force remains essentially constant during a skydive - it's what pulls the skydiver downward throughout the fall. Option B states acceleration equals gg, but this would mean the skydiver is still speeding up. At terminal velocity, acceleration is actually zero because the net force is zero. Option D incorrectly claims inertia becomes zero, but inertia is an object's resistance to changes in motion and depends only on mass, which doesn't change during the fall. For HESI physics questions, remember that "constant velocity" is a key phrase signaling force equilibrium. When you see terminal velocity or any constant-speed scenario, immediately think about balanced forces rather than the individual forces themselves. This concept appears frequently in questions about falling objects, projectile motion, and equilibrium situations.

Question 12

An object is thrown vertically upwards and takes 4 seconds to reach its maximum height. How long is its total time in the air before returning to its starting point, neglecting air resistance?

  1. 4 seconds
  2. 6 seconds
  3. 8 seconds (correct answer)
  4. 16 seconds
Explanation: When you encounter projectile motion problems involving vertical launches, remember that gravity creates perfectly symmetric motion - the upward journey mirrors the downward journey in both time and speed. Since the object takes 4 seconds to reach maximum height, you can use the symmetry principle: whatever time it takes to go up, it takes exactly the same time to come back down. This happens because gravity decelerates the object at 9.8 m/s29.8 \text{ m/s}^2 on the way up, reducing its velocity to zero at the peak, then accelerates it at the same rate on the way down. The total flight time is simply: time up + time down = 4 seconds + 4 seconds = 8 seconds. This makes C correct. Let's examine why the other options miss the mark. Option A (4 seconds) represents only half the journey - either just the upward trip or just the downward trip, but not the complete round trip. Option B (6 seconds) might tempt students who incorrectly think the downward journey takes less time than the upward journey, perhaps confusing this with air resistance scenarios. Option D (16 seconds) suggests squaring the upward time (42=164^2 = 16), which has no basis in physics and represents a mathematical error rather than proper kinematic analysis. For HESI physics problems, always look for symmetry in vertical projectile motion when air resistance is neglected. The key insight is that "what goes up must come down" - and it takes exactly the same time for both legs of the journey.

Question 13

A medical student tosses a ball straight up into the air. At the very peak of its trajectory, what are the ball's instantaneous velocity and acceleration?

  1. The velocity is zero, and the acceleration is zero.
  2. The velocity is at its maximum value, and the acceleration is directed upwards.
  3. The velocity is zero, and the acceleration is directed downwards at approximately 9.8 m/s29.8 \ m/s^2. (correct answer)
  4. Both the velocity and the acceleration are at their maximum values and are directed downwards.
Explanation: When analyzing projectile motion, you need to distinguish between velocity and acceleration at different points in the trajectory. Velocity describes how fast and in what direction an object is moving, while acceleration describes how velocity is changing. At the peak of the ball's trajectory, the ball momentarily stops moving upward before beginning to fall downward. This means its instantaneous velocity is zero - it's neither moving up nor down at that exact moment. However, gravity continues to act on the ball throughout its entire flight, constantly pulling it downward with an acceleration of approximately 9.8 m/s29.8 \ m/s^2. This gravitational acceleration never stops, even when the ball's velocity is zero. Option A incorrectly suggests acceleration is zero at the peak. Gravity doesn't "turn off" just because the ball stops moving momentarily. Option B makes two errors: velocity isn't at maximum (it's actually zero), and acceleration isn't upward - gravity always pulls downward. Option D wrongly claims both velocity and acceleration are at maximum values, but velocity is zero at the peak, not maximum. The correct answer is C because it recognizes that velocity equals zero while acceleration remains constant at 9.8 m/s29.8 \ m/s^2 downward due to gravity. Remember this key principle: in projectile motion problems, gravitational acceleration is always present and always points downward, regardless of the object's velocity. Don't confuse the moment when velocity changes direction (becoming zero) with acceleration changing - they're independent quantities.

Question 14

An astronaut has a mass of 70 kg on Earth. When she travels to the Moon, where the acceleration due to gravity is about one-sixth that of Earth, which statement is correct?

  1. Her mass will be approximately 11.7 kg, and her weight will be the same as on Earth.
  2. Her mass will be 70 kg, and her weight will be approximately one-sixth of her weight on Earth. (correct answer)
  3. Her mass will be 70 kg, and her weight will also be 70 kg, measured in different units.
  4. Her mass and weight will both be approximately one-sixth of their respective values on Earth.
Explanation: When you encounter questions about mass and weight in different gravitational environments, remember that mass and weight are fundamentally different properties. Mass measures the amount of matter in an object and remains constant regardless of location, while weight is the gravitational force acting on that mass and varies with gravitational strength. The astronaut's mass of 70 kg represents the actual amount of matter in her body. This quantity never changes whether she's on Earth, the Moon, or floating in space. However, her weight depends on the gravitational pull she experiences. On Earth, weight equals mass times Earth's gravitational acceleration (W=mgW = mg). On the Moon, where gravity is one-sixth as strong, her weight becomes WMoon=m×gEarth6W_{Moon} = m \times \frac{g_{Earth}}{6}, making her weight one-sixth of her Earth weight while her mass stays 70 kg. Option A incorrectly suggests her mass changes to 11.7 kg while weight stays the same - this reverses the actual relationship. Option C confuses mass and weight as equivalent values in different units, which ignores their distinct physical meanings. Option D incorrectly states that both mass and weight decrease by one-sixth, failing to recognize that mass is invariant. The correct answer is B because it properly distinguishes between mass (constant at 70 kg) and weight (reduced to one-sixth on the Moon). Study tip: Remember the distinction with this phrase: "Mass is stuff, weight is pull." Mass never changes, but weight depends entirely on local gravity. This concept appears frequently on science exams when discussing planetary conditions or space travel.