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

0.25 second

Step-by-step explanation:

Given :

Velocity, v = 10 m/s

The height,

`$ h = (v^2)/(2g) $`

`$ h = (10^2)/(2* 9.8) $`

= 5 m

Therefore the time at which both the bodies meet is

`$ t = (h)/(v+v) $`

`$ t = (5)/(10+10) $`

`$ t = (5)/(20) = 0.25$`

So, time taken is 0.25 seconds