A playground is on the flat roof of a city school, 4.8 m above the street below
ID: 1659636 • Letter: A
Question
A playground is on the flat roof of a city school, 4.8 m above the street below (see figure). The vertical wall of the building is h = 6.30 m high, forming a 1.5-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 = 53.0° above the horizontal at a point d = 24.0 m 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. (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.
Explanation / Answer
a) let the speed of launch be v
Horizontal velocity =vcos53
Horizontal distance travelled by ball in 2.2 seconds =24m
Therefore, vcos53*2.2 = 24
v=18.13 m/s
b)vertical velocity of ball when launched =vsin53=14.48 m/s
maximum height reached by ball=14.48²/(2*9.8)=10.7m
Time taken to reach maximum height=14.48/9.8 =1.48seconds
Since this time is less than 2.2 seconds, the ball is falling downwards while clearing the wall
Height by which the ball falls in remaining (2.2-1.48 =0.72 second) =(0.5*9.8*0.72*0.72)=2.54 m
Height of ball from ground after 2.2 seconds=10.7-2.54=8.16m
Vertical Height by which the ball clears the wall=8.16-6.3=1.86m
c)height by which the ball has to fall from maximum height to reach roof floor=10.7-4.8=5.9 m
Vertical velocity of ball just before reaching roof=sqrt(2*9.8*5.9)=10.75m/s
Time taken by ball to fall =10.75/9.8=1.1 seconds
Total time taken by ball from the time of launch to reach floor=1.48+1.1=2.58 seconds
Total horizontal distance travelled=18.13cos53 * 2.58=28.15m
Horizontal distance from roof wall where the ball lands=28.15-24=4.15 m
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