Question

A body is thrown up with a velocity 40 metre per second at the same time another body is dropped from height 40 metre the magnitude of relative velocity after t = 1 second is

20 m/s
60m/s
40m/s
10m/s
​

Answer 1

Given:

A body is thrown up with a velocity 40 metre per second at the same time another body is dropped from height 40 metre.

To find:

Relative velocity at t = 1 sec ?

Calculation:

Velocity of 1st ball at t = 1 sec is :

$\sf \therefore \: v1 = u - gt$

$\sf \implies \: v1 = 40 - (10)(1)$

$\sf \implies \: v1 = 40 -10$

$\sf \implies \: v1 = 30 \: m/s$

Now, velocity of 2nd ball at t = 1 sec is :

$\sf \therefore \: v2 = u - gt$

$\sf \implies\: v2 = 0 - (10)(1)$

$\sf \implies\: v2 = - 10 \: m/s$

So, relative velocity is :

$\sf \therefore \: \Delta v = v1 - v2$

$\sf \implies \: \Delta v = 30 - ( - 10)$

$\sf \implies \: \Delta v = 40 \: m/s$

So, relative velocity between the balls is 40 m/s.