Question

A body is dropped from a height 'h' calculate the time taken to reach the ground, the velocity of reaching the ground? ​

Answer 1

Explanation:

v = u +2aS

where S is the height...

v is final velocity...

u is initial velocity...

a is acceleration...

Answer 2

We are given:-

$s =h$

$a = 10\frac{m}{s^{2} }$ ( for easy calculations)

$u=0\frac{m}{s}$

What we need to find:-

Time and final velocity

What we need to do:-

Let us use first equation of motion.

$v=u+at$

Let us now plug in the values.

$v=0+10t$

By simplifying, we get

$v=10t$ ------- 1

Now, let us use second equation of motion

$s=ut+\frac{1}{2} at^{2}$

Let us now plug in the values.

$h=0t+\frac{1}{2} (10)t^{2}$

By simplifying, we get

$\sqrt{\frac{h}{5} } =t$  ------- 2

solving 1 and 2, we get

$v=10\sqrt{\frac{h}{5} }$

∴  $t=\sqrt{\frac{h}{5} }$  and $v=10\sqrt{\frac{h}{5} }$ .