a ball is released from a to of tower of height h.it takes time T to reach ground .height of the ball in time T/2 is nh/4 .find n
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C
8h/9 metres from the ground
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)gT
2
h=(1/2)gT
2
------[1]
in T/3 sec
h
1
=(1/2)gT
2
=(1/2)g(
3
T
)
2
=(1/2)g(
9
T
2
) -------[2]
from
[1] and [2] we get h
1
=
9
h
distance from point of release.
therefore distance from ground is h−
9
h
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