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

u= 0

a = 9.8 m/sec square

s = 19.6 m

USING THIRD EQUATION OF MOTION

v square = u square + 2as

v square = 0 + 2 × 9.8 × 19.6

v square = 384.16

v = √384.16

v = 19.6 m/sec

ANSWER = 19.6 m/sec

Answer 2

HEY DEAR ... ✌️

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Here's , Your Answer ...

=) Given , Height of tower = 19.6 m .

We know ,
Initial velocity of stone (u) = 0

Also , acceleration due to gravity (g) = 9.8 ms^-2

To find :- Final velocity of stone (v) = ?

Here ,
For free falling object =

${v}^{2} - {u}^{2} = 2gh \\ {v}^{2} - {0}^{2} = 2 \times 9.8 \times 19.6 \\ {v}^{2} = 19.6 \times 19.6 \\ {v}^{2} = 384.16 \\ v = \sqrt{384.16} \\ v = 19.6 \: {ms}^{- 1}$
Hence , the velocity of stone before touching the ground is 19.6 ms^-1 .

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HOPE , IT HELPS ... ✌️