Question

A ball is projected upwards from the foot of a tower. It crosses the top of the tower twice at an interval of 4s and reaches the ground after 12s from the start. The height of the tower is

Answer 1

135m

Explanation:

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Answer 2

Let height of tower be h.

As given,

tBCD=6sec

and tABCDE=12sec

∴tAB=3sec=tDC

and , ∴tBC=3sec=tCD

So, from relation v=u+at

From A to C,u=U

v=0,a=−g,t=6sec

O=u−gt,t=gu, or u=10×6m/s

u=60m/s

Now, from relation s=ut+21at2

from, A to B,s=h,u=60m/s,t=3sec,a=−g

h=60×3−21×10×9

h=135m