Question

A ball is released from the top of a tower of height h meters. It takes T second to reach the ground. What is the position of the ball in T/3 second - ​

Answer 1

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

So, the correct answer is “Option C”.