a body is dropped from a certain height above the ground. Its time of descent is 5s. but at t=3s, the body is stopped and then released . What is the remaining time the body should travel to reach the ground
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height = ut+1/2gt^2
=1/2×10×25
=125
height travelled in 3sec= ut+1/2gt^2
=0+1/2×10×9
=45
height left to travel=125-45
=80
as body is stopped and then again released so again u=0
h=ut+1/2gt^2
80=0+1/2×10×t^2
t^2=16
t=4sec
I hope it will help
=1/2×10×25
=125
height travelled in 3sec= ut+1/2gt^2
=0+1/2×10×9
=45
height left to travel=125-45
=80
as body is stopped and then again released so again u=0
h=ut+1/2gt^2
80=0+1/2×10×t^2
t^2=16
t=4sec
I hope it will help
Answered by
1
Answer:
Nice question, it is from Pearson Class 9.
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