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A ball dropped from a height of $1 m$ . Which of the following is the time taken by the ball to reach the ground? (Given acceleration due to gravity is $9.8$ $m s^{- 2}$ )

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0.451 s

0.0451

s

4.51

s

45.1

s

answer is A.

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Detailed Solution

The time taken by the ball to reach the ground under the given conditions is

0.451 s

.
Height at which the ball is dropped is

h = 1 m

.
Ball is dropped under the influence of gravity only so the 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 the time taken by the ball to reach the ground.
From the second equation of motion
we know that

S = ut + \frac{1}{2} a t^{2} . . . . . . . . . . . . . (a)

S

is the total distance travelled

u

is the initial velocity

t

is the time taken

a

is the acceleration
From (a) we can write this equation as

h = ut + \frac{1}{2} g t^{2} . . . . . . . . . . . . (i)

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

1 = 0 \times t + \frac{1}{2} \times 9.8 \times t^{2}

\Rightarrow 1 = \frac{1}{2} \times 9.8 \times t^{2}

\Rightarrow 1 = \frac{1}{2} \times 9.8 \times t^{2}

\Rightarrow 1 = 4.9 \times t^{2}

\Rightarrow \frac{1}{4.9} = t^{2}

\Rightarrow t^{2} = 0.204

\Rightarrow t = 0.451 s

Thus the time taken by the ball to reach the ground under the given conditions is

0.451 s

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