Question

A body is projected vertically upwards with a velocity of 10m/s from a vertically upward position at height h another body is dropped down at the same instant

Answer 1

Answer:

0.25 second  

Explanation:

Given :

Velocity, v = 10 m/s

The height,

$h = \frac{v^2}{2g}$

$h = \frac{10^2}{2\times 9.8}$

   = 5 m

Therefore the time at which both the bodies meet is

$t = \frac{h}{v+v}$

$t = \frac{5}{10+10}$

$t = \frac{5}{20} = 0.25$

So, time taken is 0.25 seconds