Question

A ball is drop from a edge of a roof it takes 0.1sec to cross a window of height 2.0m the height of the roof above the top of the window

Answer 1

We first consider the movement of ball as it covers the length of the window...

we know that

s2 = ut + (1/2)at2

now,

let the initial velocity be v1

time taken, t = 0.1s

distance, s2 = 2m

acceleration, a = g = 9.81 m/s2

so,

2 = 0.1v1 + (1/2)x9.81x0.12

or solving further

0.1v1 = 2 - 0.049

thus, we have

v1 = 19.5 m/s

now,

let us consider the situation when the ball crossed the distance between the edge of the roof and top of the window.

we know that

v2 - u2 = 2as

here,

s is to be determined

u = initial velocity = 0

v = v1 = 19.5 m/s

so,

s1 = [v2 - u2] / 2a

= [19.52 - 0] / 2x9.81

thus, we get

s1 = 19.4 m

Answer 2

Hey
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