Question

Two bodies are projected simultaneously with
the same velocity of 19.6 m/s from the top of
a tower, one vertically upwards and the other
vertically downwards. As they reach the
ground, the time gap is

​

Answer 1

Answer:

As they reach the  ground, the time gap is 4s

Explanation:

According to the problem,

Among two bodies one body which one is projected vertically upwards with a velocity of  19.6m/s  from the top will go upwards till its velocity becomes zero due to gravity, g, is acting downwards.

During this journey the time it takes:

v = u −gt;  

0 = 19.6 − 9.8t;  [ where u= 19.6m/s and g = 9.81 m/s^2]

t = 2 s

The distance it covers while going upwards within this time interval is:

h = ut−1/2gt^2

= 19.6×2 − 12×9.8×4 = 19.6mh [ where  u= 19.6m/s and g = 9.81 m/s^2 , t= 2 s]

From this point when it comes down by 19.6m;

its velocity is:

v^2 = u^2 + 2gh

      = 0 + 2×9.8×19.6 [h= 19.6m/s and g = 9.81 m/s^2 u= 0]

⇒v = 19.6m/s

The time it takes in this journey is

v = u +gt'

19.6 = 0 + 9.8t'

t' = 2 s

The procedure is same for the other body which is projected vertically downwards.

Hence t+t'= 4s