A playground is on the flat roof of a city school, h b = 5.40 m above the street

Question

A playground is on the flat roof of a city school, hb = 5.40 m above the street below (see figure). The vertical wall of the building is h = 6.50 m high, to form 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 = 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.
m/s

(b) Find the vertical distance by which the ball clears the wall.
m

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

please show a step by step response

Explanation / Answer

here,

h = 6.5 m

d = 24 m

t = 2.2 s

theta = 53 degree

a)

let the speed at which the ball is launched be v

v * cos(theta) * t = d

v * cos(53) * 2.2 = 24

v = 18.13 m/s

b)

let the vertical by which the ball clears the wall be h'

( h' + h) = v*sin(theta) * t - 0.5 * g * t^2

( h' + 6.5) = 18.13 * sin(53) * 2.2 - 0.5 * 9.8 * 2.2^2

h' = 1.64 m

c)

y = x * tan(theta) - g * x^2 /( 2 * v^2 * cos^2(theta))

6.5 = x * tan(53) - 9.8 * x^2 /( 2 * 18.13^2 * cos^2(53))

solving for x

x = 26.22 m

the horizontal distance from the wall to the point on the roof where the ball lands , x' = x - d

x' = 26.22 - 24= 2.22 m

the horizontal distance from the wall to the point on the roof where the ball lands is 2.22 m

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