A playground is on the flat roof of a city school, 5.7 m above the street below

Question

A playground is on the flat roof of a city school, 5.7 m above the street below (see figure). The vertical wall of the building is h = 6.80 m 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 = 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.)
_____ 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

Explanation / Answer

here,

theta = 53 degree

x = 24 m

time taken , t = 2.2 s

a)

let the initial speed be u

u * cos(theta) * t = x

u * cos(53) * 2.2 = 24

solving for u

u = 18.13 m/s

b)

using equation of trajectory

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

y = 24 * tan(53) - 9.81 * 24^2 /( 2 * 18.13^2 * cos^2(53))

y = 8.1 m

the vertical distance by which the ball clears the wall = y - 6.8 - 1.1 = 0.2 m

c)

let the horizontal distance be x'

using equation of trajectory

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

6.8 = x' * tan(53) - 9.81 * x'^2 /( 2 * 18.13^2 * cos^2(53))

x' = 25.8 m

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