Question

A playground is on the flat roof of a city school, 6.2m above the street below. The vertical wall of the building is h=7.30m high, forming a 1.1-m-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
θ
=53.0 degrees above the horizontal at a point d=24.0m from the base of the building wall. The ball takes 2.20s to reach a point vertically above the wall.

a) Find the speed at which the ball was launched. (Give your answer to two decimal places to reduce rounding errors in later parts.)
b) Find the vertical distance by which the ball clears the wall.
c) Find the horizontal distance from the wall to the point on the roof where the ball lands.

Answer 1

Answer:

a. 18.13m/s

b. 0.84m

c. 2.4m

Explanation:

a. to find the speed at which the ball was lunched, we use the horizontal component.Since the point distance from the base of the ball is 24m and it takes 2.20 secs to reach the wall,we can say that

t=distance /speed

$speed v_{x}=vcos\alpha=24/2.2=10.9m/s\\V=10.9/cos53^{0}\\V=18.13m/s$

Hence the speed at which ball was lunched is 18.13m/s

b. from the equation

$y=v_{y}t-\frac{1}{2}gt^{2}\\ v_{y}=18.13sin53=14.48m/s\\\\y=(14.48*2.2)-\frac{1}{2}9.8*2.2^{2}\\y=8.14m$

the vertical distance at which the ball clears the wall is

y=8.14-7.3=0.84m

c. the time it takes the ball to reach the 6.2m vertically

$6.2=14.47t-4.8t^{2}\\\\14.47t-4.8t^{2}-6.2=0\\\using\\t=-b±\frac{\sqrt{b^{2}-4ac} }{2a}\\ where \\a=-4.8, b=14.47t, c=6.2\\hence t=2.4secs$

the horizontal distance covered at this speed is

$y+24=(18.13cos53)2.4\\y=26.186-24\\y=2.12m$