Practice Problems: Projectiles Solution

1. (easy) a) Study the image below from the 2016 Rio Olympics. Compare and contrast the four paths trajectories shown.
All of the trajectories show a parabolic path, characteristic of all projectiles. The first hit (on the left) launched the volleyball with an initial velocity that had both x and y components. The second hit was launched with about the same initial x velocity, but a larger y velocity compared to the first hit. The third hit (the spike) was launched with an inital velocity only in the x direction (perhaps there is a small initial velocity in the -y direction, but it's hard to tell in the photo). The spike has a larger initial x velocity than either of the first two hits. The final hit launched the ball back up but the x directed initial velocity didn't change direction. It did, however, slow the ball down horizontally as evidenced by the distance the ball moved horizontally between successive frames. (In this analysis I assumed that the time between successive frames was constant.)

b) If the water spouts in the picture below shot the water at a slightly higher angle, would the landing place be closer to or further from the spouts? Assume Δy = 0.
Because the angle of the spouts is less than 45°, an increase in the angle will cause a greater range. The designers of this water garden must adjust the water pressure such that the exit speed of the water and the spout angle yield the desired range.

c) Analyze the snowboarder from the 2018 winter Olympics in PyeongChang. Does this athlete have a greater overall speed at takeoff or on landing? Explain your response.
Since the athlete lands at a lower elevation, the overall speed at landing is greater than the overall speed at takeoff. This is because the gravitational acceleration (only in the y-direction) has a greater time to accelerate the snowboarder on the way down than to decelerate him on the way up to max height (where v = 0).

2. (easy) Rank the range of the following projectiles:
Projectile A: Firing angle = 30°, initial speed = 40 m/s
Projectile B: Firing angle = 45°, initial speed = 40 m/s
Projectile C: Firing angle = 15°, initial speed = 40 m/s
Projectile D: Firing angle = 60°, initial speed = 40 m/s
The range of these projectiles is maximized with a 45° firing angle. Additionally, complementary angle yield the same range.
B>A=D>C

3. (moderate) A cannonball (placed on a wall 20 m above the ground) is shot at 20° firing angle with a initial speed of 17 m/s. Determine the time it takes for the cannonball to hit the ground and the distance from the base of the wall where the projectile lands. Additionally, if one assumes that the initial speed remained the same for all firing angles, what is the maximum range for the cannonball?
Assume that y_o = 20 m and x_o= 0 (positive is up and positive is to the right)
The ground is at y = 0
y - y_o = v_osinθ_o t + ½gt²
0 - 20 = 17sin(20)t - 4.9t²
0 = 20 + 5.8t - 4.9t²
t = 2.7 s
x = v_oxt = 17cos(20)(2.7)
x = 43.1 m
To find the maximum range, use a 45° initial firing angle and assume that the final height is the same as the initial height:
y - y_o = v_osinθ_o t + ½gt²
0 - 0 = 17sin(45)t - 4.9t²
0 = 12.0t - 4.9t²
t = 2.5 s
x_max = v_oxt = 17cos(45)(2.5)
x_max = 30.1 m

4. (moderate) A home run is hit in such a way as the baseball just clears a wall (21.0 m tall) located 130.0 m from home plate. The ball is hit at a 35° angle. Assume that the ball is hit 1.0 m above the ground initially. Find...
a) The initial speed of the ball
Use frame of reference wherein x_o= 0 and y_o= 1 m
x - x_o = v_ocosθ_o t
t = x/(v_ocosθ_o) = 130/(v_ocos35)
y - y_o = v_osinθ_o t + ½gt²
21 - 1 = v_osin35(130/(v_ocos35)) + (-4.9)[130/(v_ocos35)]²
v_o = 42 m/s

b) The time it takes to reach the wall
t = x/(v_ocosθ_o) = 130/(42cos35) = 3.8 s
 
c) The velocity components and the speed of the ball when it reaches the wall.
v_x = v_ocos35 = 42cos35 = 34 m/s
v_y= v_osin35 + gt = 42sin35 + (-9.8)(3.8) = -13 m/s
Thus, when the baseball clears the fence it has a velocity of v = 34i -13j for an overall magnitude of 36 m/s
v = [(34)² + (13)²]^½ = 36 m/s

5. (moderate) A projectile is launched at a 35° angle from a height of 3300 m off the ground.  It lands on the ground 9400 m (in the x-direction) from the base of the launch site.  Find the initial speed and the maximum height.
Use frame of reference wherein x_o= 0, x = 9400 m, y_o= 3300 m and y=0
x - x_o = v_ocosθ_o t
t = x/(v_ocosθ_o) = 9400/(v_ocos35)
y - y_o = v_osinθ_o t + ½gt² 
-3300 = v_osin35(9400/(v_ocos35)) + (-4.9)[9400/(v_ocos35)]²
v_o = 256 m/s
To find the max height use v_y² = v_oy² + 2aΔy
at max height v_y = 0 = (256sin35)² + 2(-9.8)Δy
Δy = 1100 m (max height is thus 4400 m from the ground)

6. (moderate)  A tennis player hits a ball with an initial speed of 23.6 m/s perfectly horizontally. The ball is 2.37 m above the ground when struck by the racquet.  If the net is 12.0 m from the ball (in the x-direction), and the net height is 0.90 m, by how much does the ball clear the net?
Use frame of reference wherein x_o= 0, x = 12.0 m, y_o= 2.37 m and y=?
x - x_o = v_ocosθ_o t
12 - 0 = 23.6(cos0)t
t =0.51
y - y_o = v_osinθ_o t + ½gt²
y - 2.37 = 23.6(sin0)(0.51) + ½(-9.8)(0.51)² 
y = 1.09 m
Since the net is 0.90 m high, the ball clears the fence by 0.19 m

7. (moderate) A rock is tossed at a 42°angle at an initial height of 1.2 m from the ground. 1.6 seconds after release, the rock reaches its maximum height. Find the initial velocity, the maximum height and the overall speed at maximum height.
At max height v_y = 0
v_y = v_oy + gt
0 = v_oy - 9.8(1.6)
v_oy = 15.7 m/s
We also know that v_oy = v_osinθ_o
15.7 = v_osin(42)
v_o = 23.5 m/s
To find max height: v_y = 0
v_y² = v_oy² + 2g(Δy)
0 = (15.7)² - 19.9(y_max - 1.2)
y_max = 13.6 m
To find overall speed at max height: v = v_x = v_ox
v = v_ocosθ_o = 23.5cos(42)
v = 17.5 m/s

8. (OPTIONAL hard) Two objects are launched from the same position, with the same initial speeds, but at different angles. The projectile fired at the lower angle bounces (upon hitting the ground) losing some of its energy while following the black trajectories. The other projectile follows the red trajectory. Find the angle θ in the sketch below.


First find the time for projectile #1:
Δy = 0 = (v_osin45)t + ½gt²
t = (v_osin45)/4.9
Find the range called D:
D = (v_ocos45)t
D = (v_ocos45)(v_osin45)/4.9 = (0.1)v_o²
t₁ = Time for projectile 2 to hit the first time.
Δy = 0 = (v_osinθ)t + ½gt²
t₁ = (v_osinθ)/4.9
t₂ = Time for projectile 2 to hit the second time.
Δy = 0 = ((v_o/2)sinθ)t + ½gt²
t₂ = (v_osinθ)/9.8
Now find D₁ and D₂
D₁ = (v_ocosθ)t₁ = v_ocosθ((v_osinθ)/4.9) = v_o²cosθsinθ/4.9
D₂ = ((v_o/2)cosθ)t₂ = ((v_o/2)cosθ(v_osinθ)/9.8) = v_o²cosθsinθ/19.6
D = D₁ + D₂
(0.1)v_o² = (v_o²cosθsinθ/4.9) + (v_o²cosθsinθ/19.6)
0.2 = (2cosθsinθ/4.9) + (2cosθsinθ/19.6)
0.2 = (sin(2θ)/4.9) + (sin(2θ)/19.6)
0.2 = 5sin(2θ)/19.6
2θ = 52°
θ = 26°

9. (moderate) A kicker on the football team gives the ball an initial velocity of 22.0 m/s.  The kick begins 40.0 m in front of the goal posts.  The goal post crossbar is 3.44 m above the ground.  Determine the minimum and the maximum kicking angles which will result in a field goal.
Clue: It might be useful for you to remember the trig identity, 1/(cos²θ) = 1 + tan²θ , as a way to simplify your analysis. Use your algebraic skills to treat the term tanθ as the unknown in a quadratic equation.
Find time with kinematic equation #2 in the x direction:
Δx = v_oxt
40 = (22cosθ)t
t = 40/(22cosθ)
Substitute the time into the y-equation:
Δy = v_oyt + ½gt²
3.44 = 22sinθ[40/(22cosθ)] - 4.9[40/(22cosθ)]²
3.44 = 40tanθ - 16.2/cos²θ
3.44 = 40tanθ - 16.2(1 + tan²θ)
0 = -16.2tan²θ + 40tanθ - 19.6  (quadratic)
tanθ = {0.67, 1.8}
θ = {34°, 61°}

10. (moderate) In every projectile example thus far, we have assumed free-fall conditions (no air resistance). Describe what you think the effect of air resistance would be on the range of a projectile. Additionally, use your ideas to predict if a projectile with an extremely big max height (very large initial speed) would have a larger range if shot at 45° or at 50°. Assume that the atmospheric density decreases with elevation.
Air resistance would cause a negative acceleration along the x-direction. Thus, the range would shorten and the trajectory would no longer be parabolic. Since the air density decreases with elevation, a projectile with a greater y-component to the initial velocity would interact with less air (overall) as it moved along its path. Thus, the 50° shot would most likely have the greater range.

Click on a link below to see Scenario Solutions:

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