A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.
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Answered by
3
v =√2gh
=√1×10×19.6
=√196
=14 m/s
=√1×10×19.6
=√196
=14 m/s
Answered by
7
_/\_Hello mate__here is your answer--
u = 0 m/s
v = ?
s = Height of the stone = 19.6 m
g = 9.8 ms−2
According to the equation of motion under gravity
v^2 − u^2 = 2gs
⇒ v^2 − 0^2 = 2 × 9.8 × 19.6
⇒ v^2 = 2 × 9.8 × 19.6 = (19.6)^2
⇒ v = 19.6 ms−1
Hence, the velocity of the stone just before touching the ground is 19.6 ms^−1.
I hope, this will help you.☺
Thank you______❤
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