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A ball dropped from a height of $1 m$ . Which of the following is the velocity of the ball at which it hits the ground? (Given acceleration due to gravity is $9.8 m / s^{2}$ )

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4.43 m / s

44.3 m / s

0.443 m / s

0.0443 m / s

answer is A.

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Detailed Solution

The velocity of the ball at which it hits the ground is

4.43 m / s

Height at which the ball is dropped is

h = 1 m

Ball is dropped under the influence of gravity only so acceleration of the ball

a = g

g = 9.8 m / s^{2}

As the ball is dropped the initial velocity of the ball

u

is zero.
We are asked to find velocity of the ball at which it hits the ground
From third equation of motion, we know that

v^{2} = u^{2} + 2 aS

---(a)

S

is the distance traveled

v

is the final velocity
From (a) we can write ( In this case )

v^{2} = u^{2} + 2 ah

−−−−(i)
Substituting all the values in the equation (i) we get

v^{2} = 0^{2} + 2 \times 9.81 \times 1

v^{2} = 2 \times 9.81 \times 1

v^{2} = 19.6

v = \sqrt{19.8}

v = 4.43 m / s

Thus the velocity of the ball at which it hits the ground is

4.43 m / s

approximately.

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