A ball is released from the top of tower of height h it takes
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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
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