### All College Physics Resources

## Example Questions

### Example Question #11 : Mechanics

A rock is tossed horizontally off the edge of a cliff with a velocity of . How long will it take to reach a total speed of ?

**Possible Answers:**

**Correct answer:**

This question requires an understanding of motion in two dimensions. The most important concept in this question is that the motion in each dimension is independent. Since the rock's initial velocity is purely in the horizontal direction, the initial velocity has no impact on the vertical velocity at any point. Likewise, the vertical acceleration has no impact on the horizontal speed. The rock will travel at in the horizontal direction throughout its entire trajectory.

To solve this problem, the first step is to set up an expression for the horizontal and vertical velocities as a function of time:

Thus,

Substituting for and solving for yields .

### Example Question #2 : Motion In Two Dimensions

A projectile reaches it max height in It has a horizontal velocity of . What is the speed at which it is launched?

**Possible Answers:**

**Correct answer:**

We first must find the initial vertical velocity () using:

We know that , since in 2 dimensions the vertical velocity at its max height is equal to zero.

we also know that because that information is given, and that

So plugging in what we know:

Knowing that is constant and equals , we can use the pythagorean theorem to determine the Resultant initial Velocity:

### Example Question #3 : Motion In Two Dimensions

If an object strikes the ground at what height was it dropped from?

**Possible Answers:**

**Correct answer:**

First let's make a table of what we know:

because it has zero velocity right when it is dropped.

It should also be noted it is simpler to define things so that the initial height is zero, so we don't have to deal with a bunch of negative numbers.

Also, since it is dropped vertically and we are only interested in the height it is dropped from we aren't interested in any information regarding the projectile's horizontal motion.

Let's use the equation:

since

We can solve for , then plug in what we know:

This is our final answer.

### Example Question #4 : Motion In Two Dimensions

A projectile is launched vertically upwards at . At what time does it reach its max height?

**Possible Answers:**

**Correct answer:**

First let's write down the information we're given:

, since at a projectile's max height its vertical velocity equals zero.

Let's use the equation

to solve for the time it takes the projectile to reach its max height.

Solving for :

Now lets plug in what we know and calculate the time:

, which is our final answer.

### Example Question #5 : Motion In Two Dimensions

A projectile is launched vertically upwards at . How long will it be in the air?

**Possible Answers:**

**Correct answer:**

First let's write down the information we're given:

, since at a projectile's max height its vertical velocity equals zero.

Let's use the following equation to solve for the time it takes the projectile to reach its max height:

Solving for :

Now lets plug in what we know and calculate the time:

Since the projectile takes twice as long to land as its does to reach its max height we simply multiple the we found by , which gives us .

### Example Question #6 : Motion In Two Dimensions

If a particle reaches its max height in , what is its range if it is launched at a speed of that remains constant throughout its flight at angle of ?

**Possible Answers:**

**Correct answer:**

First let's write out the information we are given:

We know that in order to obtain the time it takes a projectile to hit the ground we just multiply by , since the time it takes to reach the max height is half of the total time the project is in the air, so

Now we can use the equation

Since we are only concerned with the particle's motion in the horizontal direction, and we know that the horizontal velocity is constant . We also know since that is the definition of the initial position.

This gives us:

Let's plug in what we have:

. This is the range and our final answer. The reason why we can just use as the horizontal velocity is because the project is launched at a ° angle.

### Example Question #7 : Motion In Two Dimensions

A cannon is being shot from the ground. You want to shoot that cannon as far as possible. At what angle should the cannon be shot?

**Possible Answers:**

60 degrees

45 degrees

30 degrees

50 degrees

90 degrees

**Correct answer:**

45 degrees

There are many explanations for this. First you could simply insert a velocity at all of these angles and see which ends up with the greatest change in horizontal distance. You also could use basic calculus and solve for the greatest theta.

### Example Question #8 : Motion In Two Dimensions

A ball is thrown off a building at a speed of and at to the horizontal. If the building is tall, how far from the base of the building will the ball land?

**Possible Answers:**

**Correct answer:**

This is a projectile motion problem. The problem states that it was thrown off the building at an angle of 30 degrees to the horizontal. So before we can find time, we need to find the horizontal and vertical parts of this velocity.

Since and

It follows that

Now, before we can find far it went, we need to find how long it was in the air. We need to solve for time.

With the equation we can find t.

X is the initial height which in this case is the tall building. V is the initial vertical velocity and

Plugging all that in and solving yields

Knowing that the time the ball is in the air is and the constant horizontal velocity is , we can plug in these known values into the simple distance formula to solve for distance.

Certified Tutor