Question

A body is falling from height h it takes 8s to reach the ground .the time it takes to cover the first one fourth of height is?

Answer 1

Here ball is falling from rest therefore u=0 m/s.

Ball takes 8s to reach ground. Let height of ball is h.

Using 2nd equation of motion: $s=ut+\frac{1}{2}a{t}^2$

$h=0\times 8+\frac{1}{2}\times \left(10\right)\times 8^2=320m$ (for g=10$m{s}^{-2}$ )

For first one fourth of height h'=$\frac{h}{4}=\frac{320}{4}=80m$

We can find time by using 2nd equation of motion:

$80=0\times t+\frac{1}{2}\times 10\times t^2$

$80=5\times t^2$

$t^2=16\Rightarrow t=4s$

Hence it takes 4s to cover the first one fourth of height .