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Detailed Solution

A stone is dropped from a

40 m

high building. The acceleration of the stone is

20 m / s^{2}

. The velocity with which the stone strikes the ground is

40 m / s

.
Given that,
Distance travelled by the stone,

s = 40 m

Acceleration of the stone

a = 20 m / s^{2}

Initial velocity of the stone

u = 0 m / s

Final velocity,

v

is unknown
From the formula,

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

v^{2} - 0 = 2 \times 40 \times 20

\Rightarrow v = \sqrt{40 \times 40}

\Rightarrow v = 40 m / s

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