an object is dropped at height of 80 m. for what time it will remain in air?explain.(g=10m/s^2)
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Height = 80m
initial velocity = 0 m/s ( dropped )
h = 1/2 × g × t^2
80 = 1/2 × 10 × t^2
t^2 = 16
t = 4s
It will remain in air till 4s.
Hope it helps!
initial velocity = 0 m/s ( dropped )
h = 1/2 × g × t^2
80 = 1/2 × 10 × t^2
t^2 = 16
t = 4s
It will remain in air till 4s.
Hope it helps!
Jhani:
pls mark it best if understood
yea
thanks a lot
Answered by
2
S= ut + 1/2 at^2
initial vel. = u = 0
so, 80 = (1/2) 10 t^2
t^2 = 16 sec
t = 4 sec...
initial vel. = u = 0
so, 80 = (1/2) 10 t^2
t^2 = 16 sec
t = 4 sec...
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