Question

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

Answer 1

Answer: 19.6 m/s

Explanation:

As height of tower is = 19.6 m

So its obvious that the displacement =s = 19.6 m

u = 0 m/s (as it released)

a = 9.8 m/$s^{2}$ (Due to g)

Using,

$v^{2} = u^{2} + 2as$

$v^{2} = 2as$ (as u = 0)

$v^{2}$ = 2 x 9.8 x 19.6

$v^{2}$ = 19.6 x 19.6

v = 19.6 m/s

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