# Projectile Review Worksheet

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## Contents

- 1 About this page
- 2 Problems
- 2.1 1. Red Elk jumps horizontally off the edge of a cliff and hits the water 1.55 seconds later, about 12.7 m from the base of the cliff. a) What height was the cliff? b) With what speed did he leave the edge? c) What is his velocity of impact in terms of vector components, and d) in terms of angle and magnitude?
- 2.2 Red Elk projects a ball with a horizontal velocity of 13.8 m/s from the top of a cliff that is 21.0 m tall. a) What time will it take the ball to strike the ground? b) Where will the ball land? c) What will be the speed of impact? d) What is the velocity in Angle Magnitude notation) and e) position (in terms of components) of the ball exactly 1.50 seconds after he throws it?
- 2.3 3. A football leaves the ground at a speed of 27.4 m/s at an angle above the horizontal of 57
^{o}. a) Draw a picture of the initial velocity vector. b) What is the horizontal velocity? c) What is the initial vertical velocity component? d) What time will the ball be in the air? e) What distance will it go in that time? f) What is the velocity in angle magnitude notation at a time of 3.80 seconds after launch? g) What is the maximum distance you could make a football go with that speed? (Use the range equation and plug in the angle which gives you the best range) - 2.4 4. A projectile leaves the ground with a speed of 34 m/s at an angle of 37
^{o}above the horizontal. a) What is the initial velocity in vector component notation? b) What time is the projectile in the air? c) What is its range? d) What is its speed at the highest point? e) What is the velocity of the projectile in vector component notation when it is on the way up at elevation 10 m? f) Speed at elevation 10 m? - 2.5 5. A motorboat can go 2.4 m/s on a river where the current is 1.8 m/s. The motorboat must go 240 m upstream, and then back. What is the speed of that boat with respect to the shore as it travels upstream? Downstream? How much times does it take to go upstream? Downstream?
- 2.6 6. The current in a river 118 m wide is 1.45 m/s, and your boat can go 3.67 m/s. What time will it take you to cross the river if you point straight across? What speed would you go if you pointed straight in to the current? What angle upstream of straight across must you point your boat to actually go straight across?

## About this page

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Table of Contents

## Problems

### 1. Red Elk jumps horizontally off the edge of a cliff and hits the water 1.55 seconds later, about 12.7 m from the base of the cliff. a) What height was the cliff? b) With what speed did he leave the edge? c) What is his velocity of impact in terms of vector components, and d) in terms of angle and magnitude?

So we fill in our H/V suvat:

H V s = 12.7 m

u =

v =

a = 0

t = 1.55 s

s =

u = 0 (cliff)

v =

a = -9.81 m/s/s

t = 1.55 s

The first part of the problem asks for the vertical displacement. Because we have the initial velocity (0 in cliff problems), the acceleration (gravity), and the time (given), we can solve for displacement by using the following formula:

s = ut + .5at

^{2}

s = (0)(1.55 s) + (.5)(9.81 m/s^{2})(1.55 s)^{2}

s = 11.7843 m =11.8 m

The next part asks for the horizontal velocity. We can find this by using our given values

s = ut + .5at

^{2}

12.7 m = (u)(1.55 s) + (.5)(0 m/s^{2})(1.55 s)^{2}u = 8.19355 m/s =8.19 m/s

We are now looking for both of the components of the impact velocity. These are, basically, our final velocities for both x and y. Since the x value never changes, we already know our x component is going to be 8.19 m/s. We do, however, have to solve for y.

v

^{2}= u^{2}+ 2as

v^{2}= 0^{2}+ 2(-9.81 m/s^{2})(11.8 m) v = 15.2156 m/s = -15.2 m/s

8.19 m/s x + -15.2 m/s y

In order to convert that into angle magnitude, we will use our x and y components as two sides of a right triangle. Our magnitude will act as the hypotenuse. In order to find the angle, we will take the inverse tangent of the y/x.

a

^{2}+ b^{2}= c^{2}

(8.19)^{2}+ (15.2)^{2}= c^{2}c = 17.266 m/s = 17.3 m/s

tan^{-1}(-15.2/8.19) = -61.7^{o}

17.3 m/s @ -61.7^{o}

### Red Elk projects a ball with a horizontal velocity of 13.8 m/s from the top of a cliff that is 21.0 m tall. a) What time will it take the ball to strike the ground? b) Where will the ball land? c) What will be the speed of impact? d) What is the velocity in Angle Magnitude notation) and e) position (in terms of components) of the ball exactly 1.50 seconds after he throws it?

H V s =

u = 13.8 m/s

v = 13.8 m/s

a = 0

t =

s = 21

u = 0 (cliff)

v =

a = -9.81 m/s/s

t =

The first question asks for the time. Even though time is on either side, both x and y, we will use the y values to solve for it.

s = ut + .5at

^{2}

21 m = (0 m/s)(t) + (.5)(-9.81 m/s^{2})(t)^{2}

t = 20.6914 s =2.07 s

Our next variable to solve for is the horizontal displacement:

s = ut + .5at

^{2}

s = (13.8 m/s)(2.07 s) + (.5)(0 m/s^{2})(2.07 s)^{2}

s = 28.566 m =28.6 m

The next part of the question asks for the final speed. We will have to first calculate both of the final velocities and then turn them into a single magnitude (just like part d in problem 1, without us needing to find the angle). For our components, however, we can see that since there is no acceleration in x, that the final velocity for x will be the same as the initial velocity (13.8 m/s). To solve for y, however:

v

^{2}= u^{2}+ 2as

v^{2}= (0 m/s)^{2}+ (2)(-9.81 m/s^{2})(21 m)

v = 20.2983 m/s y

To now solve for the magnitude using both of our components:

a

^{2}+ b^{2}= c^{2}

(13.8)^{2}+ (20.3)^{2}= c^{2}

c = 24.5451 m/s =24.5 m/s

The final part of the question asks for both the angle magnitude and position of the ball at a given time of 1.5 s. We will first solve for the angle magnitude. No matter what our time, our x velocity is always the same (13.8 m/s). We will need to solve, however, for a new y velocity.

v = u + at

v = 0 m/s + (-9.81 m/s^{2})(1.5 s)

v = -14.715 m/s

With both the x and y components, we can use the Pythagorean Theorem and inverse tangent (like part d of problem 1) to find our angle magnitude.

a

^{2}+ b^{2}= c^{2}

(13.8)^{2}+ (14.715)^{2}= c^{2}

c = 20.1735 m/s = 20.2 m/s

tan^{-1}(-14.715/13.8) = -46.8379^{o}= -46.8^{o}

20.2 m/s @ -46.8^{o}

In order to find the displacements at both x and y at 1.5 seconds, we will use the following suvat table with our newfound values.

H V s =

u = 13.8 m/s

v = 13.8 m/s

a = 0

t = 1.5 s

s =

u = 0 (cliff)

v = 14.715 m/s

a = -9.81 m/s/s

t = 1.5 s

To solve for each the x and y displacements:

For x

s = ut + .5at^{2}

s = (13.8 m/s)(1.5 s) + (.5)(0 m/s^{2})(1.5 s)^{2}

s = 20.7 m

For y

s = ut + .5at^{2}

s = (0 m/s)(1.5 s) + (.5)(-9.81 m/s^{2})(1.5 s)^{2}

s = 11.03625 m = 11.0 m

20.7 m x + 11.0 m y

### 3. A football leaves the ground at a speed of 27.4 m/s at an angle above the horizontal of 57^{o}. a) Draw a picture of the initial velocity vector. b) What is the horizontal velocity? c) What is the initial vertical velocity component? d) What time will the ball be in the air? e) What distance will it go in that time? f) What is the velocity in angle magnitude notation at a time of 3.80 seconds after launch? g) What is the maximum distance you could make a football go with that speed? (Use the range equation and plug in the angle which gives you the best range)

We are given a vector of 27.4 m/s at 57

^{o}. In order to find the horizontal and vertical velocity components, we will have to

27.4cos(57) = 14.923 m/s =

14.9 m/s

27.4sin(57) = 22.9795 m/s =23.0 m/s

Now that we have the answers to b and c, we can create a horizontal and vertical suvat table.

H V s =

u = 14.9

v = 14.9

a = 0

t = 1.5 s

s = 0

u = 23 m/s

v = -23 m/s

a = -9.81 m/s/s

t =

We know that the vertical displacement is 0 because it's going up then coming back down on the same level. Therefore the final velocity will be the same as the initial, just negative. To solve for time, we will use the vertical side of the table

v = u + at

-23 m/s = 23 m/s + (-9.81 m/s^{2})(t)

t = 4.68909 s =4.7 s

To find the horizontal distance,

s = ut + .5at

^{2}Since a = 0,

s = ut

s = (14.9 m/s)(4.7 s)

s = 70.03 m =70 m

In order to find the angle magnitude at 3.80 s, we will first need to find the x and y components at that time. We know that since our x acceleration is always 0, our uniform x rate is 14.9 m/s (we got this from part b). We will have to solve for the y component, however, by doing the following:

Since acceleration is -9.81 m/s

^{2}for 3.8 s, the amount that the velocity will change can be expressed as (-9.81 m/s^{2})*(3.8 s) = 37.278 m/s. Adding the change in velocity from our initial y velocity of 23.0 m/s (which we got from solving part c). 23.0 m/s + -37.278 m/s = -14.278 m/s.

We can now solve for the angle magnitude using both the x and y components that we solved for.

a^{2}+ b^{2}= c^{2}

(14.9 m/s)^{2}+ (14.278 m/s)^{2}= c^{2}

c = 20.6366 m/s = 20.6 m/s

tan^{-1}(14.278/14.9) = 43.7788^{o}= 43.8^{o}

20.6 m/s @ 43.8^{o}

Using the range equation with the angle 45

^{o}, we can find the maximum distance possible. 45^{o}is the best angle possible because it is the half way point between a straight horizontal and a vertical kick.

Range = (v

^{2}/ g)*(sin(2))

Range = ((27.4 m/s)^{2}/ (9.81 m/s^{2}))*(sin(2*45))

Range = 76.53007 m =76.5 m

### 4. A projectile leaves the ground with a speed of 34 m/s at an angle of 37^{o} above the horizontal. a) What is the initial velocity in vector component notation? b) What time is the projectile in the air? c) What is its range? d) What is its speed at the highest point? e) What is the velocity of the projectile in vector component notation when it is on the way up at elevation 10 m? f) Speed at elevation 10 m?

We are given a vector of 34 m/s at 37

^{o}to convert into components.

34cos(37) = 27.1536 m/s =

27.1 m/s x

34sin(37) = 20.4617 m/s =20.5 m/s y

H V s =

u = 27.1 m/s

v = 27.1 m/s

a = 0

t =

s =

u = 20.5 m/s

v = -20.5 m/s

a = -9.81 m/s/s

t =

This being the table we can start with, we can use the vertical values to calculate our next variable, time.

v = u + at

-20.5 m/s = 20.5 m/s + (-9.81 m/s^{2})(t)

t = 4.1794 s =4.18 s

It next asks for the range, which means the horizontal distance that the object moves. In order to solve that, we will use:

s = ut + 1/2at

^{2}

s = (27.1 m/s)(4.2 s) + (.5)(0 m/s^{2})(4.2 s)

s = 113.82 m =114 m

Since this is an arc problem, at the highest point, vertial velocity is equal to 0. If horizontal velocity never changes from 27.1 m/s and the vertical velocity is equal to 0, the total velocity is going to be

27.1 m/s.

For vector component notation at any point, our x velocity will be equal to 27.1 m/s. So what we need to solve for is our vertical velocity when our vertical position is equal to 10 m.

v

^{2}= u^{2}+ 2as

v^{2}= (20.5 m/s)^{2}+ (2)(-9.81 m/s^{2})(10 m) v = 14.968 m/s = 15.0 m/s

Overall, v =27.1 m/s x + 15.0 m/s yThe next question asks for the overall speed at the same elevation. In order to do so, we will convert our x and y components into a magnitude by using the Pythagorean Theorem.

a

^{2}+ b^{2}= c^{2}

(27.1 m/s)^{2}+ (15.0 m/s)^{2}= c^{2}

c = 30.9743 m/s =31 m/s

### 5. A motorboat can go 2.4 m/s on a river where the current is 1.8 m/s. The motorboat must go 240 m upstream, and then back. What is the speed of that boat with respect to the shore as it travels upstream? Downstream? How much times does it take to go upstream? Downstream?

When the boat is traveling upstream, against the current, the two velocities must be subtracted from one another.

2.4 m/s - 1.8 m/s =

.6 m/s

When the boat is traveling downstream, the opposite occurs. Adding the velocities will give you your total velocity.

2.4 m/s + 1.8 m/s =

4.2 m/s

We can solve for time when the boat is going 240 m at a velocity of .6 m/s by division.

(240 m) / (.6 m/s) =

400 s

Applying the same principle to the downstream velocity

(240 m) / (4.2 m/s) =

57 s

### 6. The current in a river 118 m wide is 1.45 m/s, and your boat can go 3.67 m/s. What time will it take you to cross the river if you point straight across? What speed would you go if you pointed straight in to the current? What angle upstream of straight across must you point your boat to actually go straight across?

Since the question asks how long it will take to go across the river, we only need to use our horizontal velocity, 3.67 m/s. The current does not matter, it only affects the vertical displacement.

(117 m) / (3.67 m/s) = 31.8801 s =

31.9 s

The speed you would go against the current is just the speeds subtracted.

3.67 m/s - 1.45 m/s =

2.22 m/s

OK, in order to go straight across without any vertical displacement, your boat will have to be going at a vertical velocity equal to that of the current, 1.45 m/s. Your boat, however, still needs to be going a total velocity of 3.67 m/s. If you were to draw a triangle that represented this, your hypotenuse would be 3.67 m/s and your vertical side would be 1.45 m/s. We can find the third side, the horizontal velocity, by using the Pythagorean Theorem.

a

^{2}+ b^{2}= c^{2}(1.45 m/s)^{2}+ b^{2}= (3.67 m/s)^{2}b = 3.37 m/s

Using our x and y components of 3.37 m/s x + 1.45 m/s y, we can use inverse tan to find the angle.

tan

^{-1}(1.45/3.37) = 23.2806^{o}=23.3^{o}