### All AP Physics 1 Resources

## Example Questions

### Example Question #11 : Motion In Two Dimensions

Let represent the magnitude of gravity acceleration.

As in the given, figure, a ball is kicked from the ground giving it an initial velocity of at an angle of . It takes a time to reach the highest point in its trajectory. When it finally reaches the ground, which is perfectly flat, the ball covers a range (the total displacement on the horizontal axis) of . Which of the following equations represents the range, , in terms of , , , and ?

**Possible Answers:**

**Correct answer:**

An important thing to notice is that this projectile motion is symmetrical, meaning that the ball starts and ends its motion at the same height. Then, it must be the case that the ball takes the same amount to reach its highest point from its original position as it takes the ball to fall down from the highest point to its final position. In other words, it takes a time of to complete the entire trajectory. Since the ball's movement follows projectile motion, it experiences motion with constant velocity on the horizontal axis (the x-axis). We can calculate this velocity by obtaining the x-component of the ball's initial velocity:

Since the velocity is constant, we can express it in terms of displacement in the x-direction and time with the following equation:

Here, is the displacement in the x-direction and is the time.

Therefore, we have that for the range (the ball's entire displacement in the x-direction for its motion) is given by:

### Example Question #11 : Motion In Two Dimensions

Let represent the magnitude of the gravity acceleration.

As in the given figure, a ball is kicked from the ground giving it an initial velocity of at an angle of . It takes a time to reach the highest point in its trajectory. When it finally reaches the ground, which is perfectly flat, the ball covers a range (the total displacement on the horizontal axis) of . In terms of , , , and , how much time does it take the ball to reach its highest point?

**Possible Answers:**

None of these

**Correct answer:**

How much time it takes the ball to reach its highest point depends on its motion in the y-axis. Therefore, it depends on the ball's initial velocity in the y-direction and the acceleration of gravity. Note that at the highest point of its trajectory, the ball reaches a velocity of in the y-direction for an instant, just before it changes direction and starts coming down. Therefore we can use the following kinematic equation for motion with constant acceleration to find :

Here, is the final velocity on y for this part of the ball's trajectory (i.e. its velocity at the highest point), is the ball's initial velocity in the y-direction, and is the acceleration acting on the y-axis (gravity). We can express in terms of and by obtaining the y component of the initial velocity of the ball, which is given by:

Note that the acceleration in y-direction is negative since gravity pulls downwards.

Solve for .

### Example Question #11 : Motion In Two Dimensions

A cannon ball is fired at an angle of 45 degrees from the horizontal. Which of the following describes the kind of motion experienced by the cannon ball in the y-axis and the x-axis respectively?

**Possible Answers:**

Y-axis: motion with constant velocity and x-axis: motion with constant velocity

Y-axis: motion with constant velocity and x-axis: motion with constant acceleration

y-axis: motion with constant acceleration and x-axis: motion with constant velocity

Y-axis: motion with constant acceleration and x-axis: motion in free fall

Y-axis: motion in free fall and x-axis: motion with constant acceleration

**Correct answer:**

y-axis: motion with constant acceleration and x-axis: motion with constant velocity

The cannon ball has a two-dimensional motion, specifically projectile motion. Any two-dimensional motion consists of two one-dimensional motions happening at the same time. This is why we break vectors into components, so we can analyze each one-dimensional motion separately. The motion in the y-axis and the x-axis are connected by the time variable, since they happen at the same time. For projectile motion, we have that on the y-axis (the vertical axis) a constant acceleration (gravity) is acting on the cannon ball. For the y-axis motion with constant acceleration or free fall (which is motion with constant acceleration where the acceleration is gravity) are correct choices. On the x-axis (the horizontal axis), there is no acceleration acting on the cannon ball horizontally. Therefore, on the x-axis the cannon ball experiences motion with constant velocity. Then the only answer choice that describes the motion of the cannon ball in each axis correctly is: y-axis: motion with constant acceleration and x-axis: motion with constant velocity.

### Example Question #381 : Ap Physics 1

An object weighing 15kg is fired from 30m above the ground at an angle of to the horizontal with an initial velocity of . Determine how long it takes the object to reach its maximum height. Ignore friction.

**Possible Answers:**

**Correct answer:**

At the projectile's maximum height, we know that it's vertical velocity will be Since the only type of acceleration the object faces is the acceleration due to gravity, we can model its velocity using only two terms: it's initial velocity and acceleration due to gravity. Recall that the formula for velocity with constant acceleration is given as:

Here, is the object's constant acceleration, is the initial velocity, and is time.

The only acceleration is acceleration due to gravity, which is:

To determine initial velocity , we have to determine a few other things. The object is being launched at a speed of at an angle of . Therefore, the vertical component (the vertical velocity of this object) is going to be

Therefore, we can model the velocity of this object as:

The signs of the two terms must be opposite because they are going in different directions. Negative velocities in this model mean the object is moving towards the center of the earth.

Since the object reaches its maximum point when we must solve the following equation for

### Example Question #13 : Motion In Two Dimensions

A 2 kg cat is stuck in a 10 meter high tree. Just as a fire fighter atttempts to save the cat, the cat slips and falls out of the tree. Luckily, Superman is standing only 3 kilometers away and sees the cat fall. Assuming Super man has instananeous reactions and begins running towards the cat just as it falls, how fast does Superman have to run in order to catch the cat before it hits the ground?

**Possible Answers:**

**Correct answer:**

Superman needs to cover 3,000 meters. But the question is how much time does he have to cover those 3,000 meters? If you know the time, you can just divide 3,000 meters by the time he is traveling and you have his speed! But how do we know how quickly he needs to cover the 3,000 meters? The falling cat tells us! The acceleration due to gravity is a constant on the surface of Earth and it will dictate how quickly something hits the ground once dropped. Considering we ignore air resistance in physics 1, we don't care about the object's size or mass. So all we need to do is find the appropriate kinematic equation(these equations relate displacement, velocity, acceleration and time) and solve for time. The way I chose a kinematic equation is I organize variables into things we are told in the problem and things we are not told in the problem. In this case, we are given information about the height of the cat, it's mass and that it slipped (velocity in both x and y directions is zero). Not much to go off of but it is all we need! We are solving for time and we have information about initial velocity and displacement.

The equation:

We can see with this equation that under the assumption that the initial velocity is zero, it simplifies to a relationship between x and t (acceleration is g and is a constant). Rearranging for t gives:

Plugging in the appropriate numbers, the time that it take for the cat to hit the ground will be 1.43 seconds. Now we know Superman must cover the 3,000 in 1.43 seconds.

Dividing 3,000 by 1.43 yelids a velocity of

.

### Example Question #14 : Motion In Two Dimensions

A cannon is fired from the top of a cliff as shown in the figure below. If the cannonball travels a horizontal distance of in , at what angle was the cannon fired? (Hint: Find an expression for . You do not need the initial velocity to solve this.)

**Possible Answers:**

**Correct answer:**

At first this problem could seem challenging, but in fact, it is straight-forward if you can remove the initial velocity from the equations. We begin by writing the equations of motion for the x and y directions.

We then devide the two equations to get an expression for the tangent of the angle, as well ass remove the initial velocity from the problem. This enables us to use the information given to calculate the angle the projectile was fired.

### Example Question #12 : Motion In Two Dimensions

Determine how long it takes a car to move up a incline, with its top at above the ground if its speed is constant at .

**Possible Answers:**

**Correct answer:**

The top of the hill is above the ground. However, that isn't the distance from the bottom of the hill to the top, since the car is going up an incline of .

The length of the incline can be given as:

, where is the length of the incline.

Solving for

.

The incline is in length.

At a speed of ,

### Example Question #15 : Motion In Two Dimensions

A projectile is shot at above the horizontal at a velocity of . How high does the projectile travel at its peak?

**Possible Answers:**

**Correct answer:**

First we need to find the vertical velocity component (y-component). We know our velocity is at an angle of . We need to use the sine function to determine the y-component:

Now we use the kinematic equation to determine the height reached:

We solve for which is our height and substitute our known values.

Our initial vertical velocity is and final vertical velocity is at the peak.

### Example Question #13 : Motion In Two Dimensions

A projectile from a hand gun is fired at above the horizontal at a speed of . How many seconds is the projectile airborne? Assume and that the hand gun was fired a negligible distance from the ground.

**Possible Answers:**

None of these

**Correct answer:**

Since the question is asking how long the projectile was airborne, we are dealing with motion in y-direction only. Find initial velocity in the y-direction and substitute into the kinematic equation:

### Example Question #15 : Motion In Two Dimensions

A medieval cannon fires its projectile horizontally from a castle wall 70 meters high at a speed of . How long will the cannonball be airborne?

**Possible Answers:**

**Correct answer:**

Narrow the choices of kinematic equations by including only ones with distance in their formulas. Of those two, we want the one that includes time:

The red term becomes zero because our projectile is not moving at first. Plug in known values and solve.

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