Question

A ball dropped from top of the building pass the window of buidling in 0.1 s.find the hieght of the window if distance between top of window to point where ball dropped is 4.9m.

Answer 1

given that ball is dropped from 4.9 m height

now by using kinematics equations

$s = v_i*t + \frac{1}{2}gt^2$

$4.9= 0 + 4.9 t^2$

$t = 1s$

Now its given that it cross the window in 0.1 s

now again by kinematics

$s = v_i *t + \frac{1}{2}gt^2$

$s = 0*1.1 + \frac{1}{2}*9.8*1.1^2$

$s = 5.93 m$

So now the length of window will be

$L = 5.93 - 4.9= 1.03 m$

So height of window will be 1.03 meter