1.

A stone is released from the from the top of a tower of height 19.6m calculate its final velocity just before touching the ground.

Answer»

Solution :According to the equation of motion under GRAVITY.
`v^(2)-u^(2)=2gs`
Where, u=Initial VELOCITY of the stone=o
v=Final velocity of the stone
s=Height of the stone =19.6m
g-Acceleration DUE to gravity `=9.8ms^(-2)`
Therefore `v^(-2)-o^(2)=2xx9.8xx19.6`
`V^(2)=2xx9.8xx19.6=(19.6)^(2)`
`V=19.6ms^(-1)`
Hence, the velocity of the stone just before TOUCHING the ground is `19.6ms^(-1)`.


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