Question

A ball is released from the top of the tower of height h metres. It takes T seconds to reach the ground. What is the position of the ball from the ground in T/3 seconds?

Answers with steps

Answer 1

Answer:

ANSWER

the acceleration of the ball will be g. Initial velocity will be 0. 

in T sec. body travels h mts.  

by applying equations of motion we get

s=ut+(1/2)gT2

h=(1/2)gT2       ------[1]

in T/3 sec     

  h1=(1/2)gT2=(1/2)g(3T)2=(1/2)g(9T2)     -------[2]

from

 [1] and [2] we get h1=9h distance from point of release.

therefore distance from ground is h−9h=98h

Explanation:

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