Question

A body is dropped from top of a tower.It acquires a velocity of 20ms-2 on reaching the ground .(gravity=10ms-2).Find the height of the tower.

Answer 1

Answer:

Height of the tower = 10 m

Explanation:

As per the 3rd equation of motion;

$v^2 = u^2 + 2as$

Here, v = 20 m/s

         u = 0

          a = g = 10 m/$s^2$

So, $20^2 = 0 + 2*10*h$

so, h = 400/20 m

h = 10m

Height of the tower = 10 m

Answer 2

Answer:

A body is dropped from top of tower

so initial velocity (u) = 0

and final velocity (v)= 20 m/s

FROM THIRD EQUATION OF MOTION

V^2 - u^2 = 2gh ( g is gravity )

20^2 - 0^2 = 2×10×h

400 = 20h

h = 400/20

h = 20 m

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