A ball is dropped from height ‘h’ in the last second it
travels 9h/25. Find h.
1) (25/2) g 2) (15/2) g
3) (5/2) g 4) (35/2)g
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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
Explanation:
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