Question

A body released from the top of a tower of height h takes T seconds to reach the ground . The position of the body at T/4 second is...

Answer 1

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 +1/2gT^2
h = 1/2gT2 ------[1]
in T/4 sec h1 = 1/2gT^2/16 -------[2]
from [1] and [2] we get h1 =h/16 distance from point of release.
therefore distance from ground is h-h/16
=15h/16

Answer 2

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 +1/2gT^2

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

in T/4 sec h1 = 1/2gT^2/16 -------[2]

from [1] and [2] we get h1 =h/16 distance from point of release.

therefore distance from ground is h-h/16

=15h/16