Question

**3)** A ball is thrown leftward from the left edge
of a roof at height “h” above the ground. The ball hits the ground
1.5 sec later at a distance d = 25 m from the building and at an
angle θ = 60° with the horizontal.

a) What is the value of “h”?

b) What is the magnitude of the initial velocity?

c) What is the angle of the initial velocity?

Answer #1

horizontal velocity

Vx = 25 / 1.5 =16.667 m/s

net velocity just before impact

Vx = V cos 60

16.667 = V cos 60

V = 33.33 m/s

vertical component of velocity

vy = V sin 60 = 28.867 m/s

using 1st equation of motion

vy = uy + gt

uy = 28.867 - 9.8* 1.5

uy = 14.167 m/s

ux = vx = 16.667 m/s

net velocity of projection

u^2 = ux^2 + uy^2

u = 21.874 m/s ====>>>> Ans (b)

direction

x = arctan ( uy/ux) = 40.36 ===>>> Ans (c)

using 3 rd equation of motion

v^2 = u^2 + 2 g h

h = 32.266 m===>>> Ans (a)

====

Comment before rate in case any doubt, will reply for sure.. goodluck

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