Projectile Problems with Solutions 3

 Problem#1

A place-kicker must kick a football from a point 36.0 m (about 40 yards) from the goal, and half the crowd hopes the ball will clear the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 20.0 m/s at an angle of 53.0° to the horizontal. (a) By how much does the ball clear or fall short of clearing the crossbar? (b) Does the ball approach the crossbar while still rising
or while falling?

Answer:
(a) We use the trajectory equation:

y_f = y_i + v_yit + ½ a_yt² and x_f – x_i = v_xit₁

with v_xi = v_i cos θ_i and v_yi = v_y sin θ_i, then

y_f = y_i­ + x_f tan θ_i – gx_f²/(2v_i² cos θ_i)

given, x_f = 36.0 m, v_i = 20.0 m/s, and θ_i = 53.0⁰, we find

y_f = 0 + (36.0 m) tan 53.0⁰ – (9.80 m/s²)(36.0 m)²/{2 x (20.0 m/s)² cos 53.0⁰}

y_f = 3.94 m

The ball clears the bar by

(3.94 m – 3.05 m) = 0.889 m

(b) The time the ball takes to reach the maximum height (v_y = 0) is

v_y = v_yi + (–g)t_m

0 = 20.0 m/s sin 53.0⁰ – (9.80 m/s²)t_m

t_m = 1.63 s
The time to travel 36.0 m horizontally is

t = x_f/v_xi

t = 36.0 m/(20.0 m/s x cos 53.0⁰) = 2.99 s

since t > t_m, the ball clears the goal on its way down

Problem#2
A firefighter, a distance d from a burning building, directs a stream of water from a fire hose at angle 'i above the horizontal as in Figure 1. If the initial speed of the stream is vi, at what height h does the water strike the building?

Answer:
The horizontal component of displacement is

x_f = x_i + v_xit = v_i cos θ_i.t

Therefore, the time required to reach the building a distance d away is

t = x_f/v_i cos θ_i

At this time, the altitude of the water is

y_f = y_i + v_yit + ½ a_yt²
y_f = y_i­ + v_i sin θ_i (x_f/v_i cos θ_i) – ½ g (x_f/v_i cos θ_i)²

h = d tan θ_i – gd²/(2v_i² cos θ_i)

Problem#3
A playground is on the flat roof of a city school, 6.00 m above the street below. The vertical wall of the building is 7.00 m high, to form a meter-high railing around the playground. A ball has fallen to the street below, and a passerby returns it by launching it at an angle of 53.0° above the horizontal at a point 24.0 meters from the base of the building wall. The ball takes 2.20 s to reach a point vertically above the wall. (a) Find the speed at which the ball was launched. (b) Find the vertical distance by which the ball clears the wall. (c) Find the distance from the wall to the point on the roof where the ball lands.

Answer:
(a) For the horizontal motion, we have

x_f = x_i + v_xit = v_i cos θ_i.t

24.0 m = 0 + (v_i cos 53⁰)(2.2 s)

v_i = 18.1 m/s

(b) As it passes over the wall, the ball is above the street by

y_f = y_i + v_yit + ½ a_yt²

y_f = 0 + (18.1 m/s)(sin 53⁰)(2.2 s) + ½ (–9.80 m/s²)(2.2 s)

y_f = 8.13 m

So it clears the parapet by 8.13 m – 7 m = 1.13 m

(c) Note that the highest point of the ball’s trajectory is not directly above the wall. For the whole flight, we have from the trajectory equation

y_f = y_i­ + x_f tan θ_i – gx_f²/(2v_i² cos θ_i)

or

6.0 m = x_f(tan 53⁰) – [(9.80 m/s²)/2(18.1 m/s)² cos² 53²] x_f²

simplified to be

0.0412x_f² – 1.33x_f + 6 m = 0

x_f = {1.33 ± √[1.33² – 4(0.0412)(6)]^1/2}/{2(0.0412)}

x_f = 26.8 m or x_f = 5.44 m

The ball passes twice through the level of the roof. It hits the roof at distance from the wall

26.8 m – 24.0 m = 2.79 m

Problem#4
A dive bomber has a velocity of 280 m/s at an angle θ below the horizontal. When the altitude of the aircraft is 2.15 km, it releases a bomb, which subsequently hits a target on the ground. The magnitude of the displacement from the point of release of the bomb to the target is 3.25 km. Find the angle θ.

Answer:

When the bomb has fallen a vertical distance 2.15 km, it has traveled a horizontal distance x_f given by (Fig.3)

x_f = √[(3.25 km)² – (2.15 km)²] = 2.437 km

and

y_f = y_i­ + x_f tan θ_i – gx_f²/(2v_i² cos θ_i)

0 = 2150 m + 2437 m (tan θ) – [(9.8 m/s²)(2437 m)²]/[2(280 m/s)²cos²θ]

–2150 m = 2437 m (tan θ) – (371.19 m)(1 + tan²θ)

Note: sec² θ = 1+ tan² θ

tan²θ – 6.565 tanθ – 4.795 = 0

tan θ = ½ [6.565 ± √{(6.565)² – 4(–4.795)}] = 3.283 ± 3.945

Select the negative solution, since θ is below the horizontal, then

tan θ = –0.662 or θ = –33.5⁰.