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A boy standing at the top of a tower of 20 m height drops a stone. Assuming g = 10 m $s^{-2}$, the velocity with which it hits the ground is
A
5.0 m/s
B
10.0 m/s
C
20.0 m/s
D
40.0 m/s
Detailed Solution
The stone is dropped, so initial velocity u = 0; height fallen h = 20 m; g = 10 m $s^{-2}$.
Using $v^2 = u^2 + 2gh$
$v^2 = 0 + 2\times10\times20 = 400$
$v = \sqrt{400} = 20$ m/s
Using $v^2 = u^2 + 2gh$
$v^2 = 0 + 2\times10\times20 = 400$
$v = \sqrt{400} = 20$ m/s