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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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20 m / s

19.6 m / s

15.3 m / s

12.3 m / s

answer is B.

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Detailed Solution

A stone is released from the top of a tower of height

19.6 m

. The velocity of the ball just before touching the ground is

15.3 m / s

.
Given data in the problem,
Initial velocity,

u = 0

Height of tower,

s = 19.6 m

By using the third equation of motion,

v^{2} - u^{2} = 2 as

Where,

u = initial velocity

v = final velocity

a = acceleration

s = distance

Here acceleration is due to gravity. So,

a = g

\Rightarrow v^{2} = u^{2} + 2 gs

\Rightarrow v^{2} = 0 + 2 \times 9.8 \times 19.6

\Rightarrow v^{2} = 384.16

\Rightarrow v = \sqrt{384.16}

\Rightarrow v = 19.6 m / s

The velocity before touching the ground is

19.6 m / s

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