A ball is released from the top of a tower of height h metres. It takes 'T' seconds to reach the ground. What is the position of the ball in T/3 seconds ?
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5T/3will be the distance
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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)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
=
9
8h
make me briliant
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