Question

a body is released from the top of height h. it takes t sec to reach the ground. where will be the ball after time t/2

Answer 1

Answer:

Explanation:

The acceleration of the ball will be g.

Initial velocity will be 0.

In T sec the body travels h.  

by applying equations of motion we get

s= ut +1/2gT2

h = 1/2gT2 ------[1]

in T/3 sec h1 = 1/2gT2/9 -------[2]

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

therefore distance from ground is h-h/9 =8h/9.

Answer 2

Answer:

please refer to the the attachment and Mark it as brainliest if it helps you