Question

a body released from top of a tower of height h takes T seconds to reach the ground then the position of the body at T/4 seconds is.
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Answer 1

Answer:

apna app kiya karo mam na padaya nhi kya

thanks

Answer 2

Explanation:

The acceleration of the ball is g ,

Its initial velocity U=0,

We know that distance s is ,

s=ut+12at2

But u=0, and acceleration is equal to g so,

or h=12gT2 (U=0)

Therefore, in T/3 seconds,

h′=12g(T3)2

On solving the equation we get,

⇒h′=12×gT29=h9

Now subtracting the distance travelled in T/3 seconds from the total distance of the tower,

=h−h9=8h9