Answer to Question #162513 in Physics for Chrishan

Question

A playground is on the flat roof of a city school, 7.50 m 

above the street below. The vertical wall of the building is 

h = 9.00 m high, to form a 1.50-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 = 54.5° 

above the horizontal at a point d = 25.0 m from the base of 

the building wall. The ball takes 2.05 s to reach a point 

vertically above the wall. (a) Find the speed at which the 

ball was launched. (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

a)

$$25=2.05\cos{54.5}v_i\\v_i=21.0\frac{m}{s}$$

b)

$$y_f=(21)(2.05)\sin{54.5}-0.5(9.8)(2.05)^2=35\ m$$

$$s=35-9=26\ m$$

c)

$$7.5=\tan{54.5}x_f-\left(\frac{9.8}{2(21)^2\cos^2{54.5}}\right)x_f^2\\x_f=6.28\ m\ or\\36.3\ m$$