A playground is on the flat roof of a city school, h_b = 6.10 m a the street bel

Question

A playground is on the flat roof of a city school, h_b = 6.10 m a the street below (see figure). The vertical wall of the building is h = 7.60 m high, to form a 1.5-m-high railing around the above playground. A ball has fallen to the street below, and a passerby returns it by launching it at an angle of theta = 53.0 degree 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. Find the speed at which the ball was launched. Find the vertical distance by which the ball clears the wall. Find the horizontal distance from the wall to the point on the roof where the ball lands.

Explanation / Answer

Given data
ho = 6.10m
h= .60m
height of the railing =1m
theta = 53 degrees
time = 2.20s
d = 24 m

a) Find the speed at which the ball was launched.
2.2*cos(53)Vo = 24
Vo = 18.1m/s

S= Vo*t-1/2*g*t^2
=2.2*sin(53)(18.1) - 0.5*9.8*(2.2)^2

Assuming the passerby has an initial height of 0.
The ball is 8 feet up, which means it is 8 - 7.60 = 0.40 meter above the wall.

b) Find the vertical distance by which the ball clears the wall.
1.5 meterhigj railing
New equation is now
6.10 = sin(53)(t)(18.1) - 0.5*9.8*(t)^2
There will be two times, you want the larger time. This time is 2.439 seconds
which you will plug in to the first eq
cos(53)18.1*2.439 = d
d = 26.567 m
d - 24 = 2.567 m

c) Find the horizontal distance from the wall to the point on the roof where the ball lands.

the distnce is = 2.567 m

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