The Physics
Hypertextbook
Opus in profectus

# Projectiles

## Practice

### practice problem 1

In the 1994 action-adventure film Speed, an extortionist equipped a Los Angeles bus with a bomb that was set explode if the speed of the bus fell below 50 mph (22 m/s). The police discovered the bomb and routed the bus on to a segment of freeway that was still under construction — their intention being to keep it out of the notoriously heavy Southern California traffic. In a twist of the plot, however, they overlooked a 50 foot (15 m) gap in the construction. Since the bomb would explode killing everyone on board if they slowed down, they decide to jump the gap. The missing segment lies on an effectively level stretch of elevated freeway, 50 feet (15 m) above the ground, yet somehow they manage to jump the gap at 67 mph (30 m/s) and land safely on the other side.
1. If the bus in Speed obeyed the laws of physics how would this scene end?
2. Just out of curiosity, what would happen if the bus in the movie Speed was driven off of a 15 m high, horizontal ramp with nothing but the ground to land on?
1. Where would the bus land?
2. What velocity would the bus have when it struck the ground? State its…
1. magnitude and
2. direction
3. Let's modify this stunt so that a bus like the one in the movie could jump over a 50 foot (15 m) gap in the freeway.
1. How long does it take to jump the gap traveling at the speed shown in the movie?
2. What upward velocity would a bus need to stay in the air this long?
3. What direction should the bus head?
4. What type of device is needed to make the bus travel in this direction?

#### solution

1. This question is basically asking how far down a bus moving at 30 m/s would fall in the time it took it to move forward 15 m. Although the bus is accelerating downward due to gravity its horizontal velocity remains constant. (Gravity never acts horizontally.) It will take the bus…

 Δt = Δx v
 Δt = 15 m 30 m/s
 Δt = 0.50 s

to travel the width of the gap, during which time it would fall…

 y = ½at2y = ½(9.8 m/s2)(0.50 s)2y = 1.23 m

This means the bus will strike the flat concrete face of the freeway stub somewhere above its front bumper. So much for Speed II.

2. Let the bus fly free!
1. The first half of this question is basically asking how far forward a bus moving at 30 m/s would travel in the time it took for it to fall 15 m downward. In this problem there are two independent equations of motion — one with constant velocity (the horizontal motion) and one with constant acceleration (the vertical motion). Since the ramp is horizontal, there is no initial velocity in the vertical direction. Thus, the time it takes a horizontally launched projectile to reach the ground is the same as the time it takes an object released from rest to fall the same height.

 y0 = 0 m y = 15 m v0y = 0 m/s ay = 9.8 m/s2 ∆t = ?
 y = y0 + v0yΔt + ½ayΔt2 y = ½ayΔt2 Δt = √(2y/ay) Δt = √[2(15 m)/(9.8 m/s2)] Δt = 1.75 s

Now use the time of flight to determine the horizontal displacement of the bus as it flies through the air.

 vx = 30 m/s Δt = 1.75 s x = ?
 x = vxΔtx = (30 m/s)(1.75 s)x = 52.5 m

Look for the bus about 53 m in front of a point directly below the edge of the ramp.

2. The final velocity of the bus will be the vector sum of its horizontal and vertical velocities on impact. Since there is no horizontal acceleration, the velocity in this direction remains constant. If the bus leaves the ramp at 30 m/s horizontally it will strike the ground at 30 m/s horizontally. There is an acceleration vertically, however. Initially, the bus has no vertical velocity, but this will obviously change. Gravity pulls everything relentlessly toward the Earth. Once there's no more road to hold up the bus, there's nothing to keep it from accelerating down.

 y0 = 0 m y = 15 m v0y = 0 m/s ay = 9.8 m/s2 vy = ?
 vy2 = v0y2 + 2ayy vy = √(2ayy) vy = √[2(9.8 m/s2)(15 m)] vy = 17.1 m/s

Now that we have both components, find the magnitude and direction of the resultant velocity by the usual means: Pythagorean theorem…

 v2 = vx2 + vy2 v2 = (30 m/s)2 + (17.1 m/s)2 v = 35 m/s

and tangent…

tan θ =
 vy = 17.1 m/s vx 30 m/s
θ =  30° (angle of depression)

3. Let's make a movie like this one.
1. Although the bus is accelerating downward due to gravity its horizontal velocity remains constant. (Gravity never acts horizontally.) It will take the bus…

 Δt = Δx v
 Δt = 15 m 30 m/s
 Δt = 0.50 s

to travel the width of the gap.

2. If the bus needs half a second to jump up and return to the freeway, it needs a quarter of a second to reach the highest point in its jump. At this point, it is not moving vertically. Use this information to determine the initial vertical speed of the bus so that it can jump the gap.

 Δt = 0.25 s vy = 0 m/s a = −9.8 m/s2 v0y = ?
 vy = v0y + aΔt(0 m/s) = v0y + (−9.8 m/s2)(0.25 s)v0y = 2.45 m/s
3. Directions are ususally found from components using the tangent function. This problem is no different.
 tan θ = vy vx
 tan θ = 2.45 m/s 30 m/s
 tan θ = 5° above the horizontal
4. The bus needs a ramp to jump this gap. If you watch this scene in the movie carefully, you will see a small ramp appear very briefly.

### practice problem 2

Projectile launcher demo.
1. Launch horizontally from a convenient height above the floor (to a reproducible location).
1. Measure the horizontal distance traveled.
2. Measure the vertical height fallen.
3. Calculate the time to fall.
4. Calculate the horizontal velocity.
2. Launch from an interesting angle above the horizontal.
1. Measure the angle of inclination.
2. Determine the horizontal component of the projectile's initial velocity.
3. Determine the vertical component of the projectile's initial velocity.
4. Determine the time to reach the highest point.
5. Determine the height of the highest point (above the launch height).
6. Determine the location where the projectile returns to its launch height.

#### solution

The results of this demo vary from year to year. Here's a set of typical outcomes. Your results may vary.

1. Launch horizontally from a convenient height above the floor (to a reproducible location).
1. This year, the horizontal distance was…

x = 3.32 m

2. This year, the height of the launcher was…

y = 1.18 m

3. A horizontally launched projectile has no initial velocity in the vertical direction. It falls the same as an object released from rest.

 y = 1.18 m v0 = 0 m/s a = 9.8 m/s2 t = ?
 y = v0t + ½at21.18 m = ½(9.8 m/s2)t2t = 0.491 s
4. A projectile might fall like an object in free fall, but it moves forward like an object in motion that wants to stay in motion. It's horizontal velocity is constant.

 x = 3.32 m t = 0.491 s v = ?
 v = x/tv = (3.32 m)/(0.491 s)v = 6.77 m/s
2. Launch from an interesting angle above the horizontal.
1. This year, the angle of inclination was…

θ = 50°

2. The horizontal component of the projectile's initial velocity can be found using the cosine of the angle of inclination.

 vx = v cos θvx = (6.77 m/s) cos 50°vx = 4.35 m/s
3. The vertical component of the projectile's initial velocity can be found using the sine of the angle of inclination.

 vy = v sin θvy = (6.77 m/s) sin 50°vy = 5.18 m/s
4. This is a question about stopping — going from some speed to no speed — in the vertical direction. How long does that take when gravity is what's slowing you down? Rearrange the definition of acceleration. Slowing down is a negative change in velocity and a negative acceleration. Two negatives make a positive, so don't bother writing any signs.

 t = ∆vy/at = (5.18 m/s)/(9.8 m/s2)t = 0.529 s
5. This problem can be solved in one of two ways. For both methods, gravity is taken as negative since the projectile is slowing down in the vertical direction. First we'll solve it using the time calculated in the previous part.

 t = 0.529 s v0 = 5.18 m/s a = −9.8 m/s2 t = ?
 y = v0t + ½at2y = (5.18 m/s)(0.529 s) + ½(−9.8 m/s2)(0.529 s)2y = 1.37 m

Next we'll solve it without using time.

 v0 = 5.18 m/s v = 0 m/s a = −9.8 m/s2 y = ?
 v2 = v02 + 2ay0 m/s = (5.18 m/s)2 + 2(−9.8 m/s2)yy = 1.37 m
6. The projectile is not accelerating horizontally so distance is just speed times time — twice the time it takes to go halfway.

 x = vxtx = (4.35 m/s)(2 × 0.529 s)x = 4.60 m

### practice problem 3

Shoot the monkey demo

#### solution

Describe the solution.

### practice problem 4

1. max range 45°
2. max height 90°
3. max flight time 90°
4. max path length 56.46°
5. max area under curve 60°