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If a ball is dropped from a height of $20 m$ , and its velocity increases uniformly at the rate of $10 m s^{- 2}$ , then with what velocity and time will it strike the ground?

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20 m s^{- 1}, 2 s

2 m s^{- 1}, 20 s

2 m s^{- 1}, 10 s

40 m s^{- 1}, 2 s

answer is A.

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Detailed Solution

It will strike the ground with

20 m s^{- 1}

velocity and

2 s

time.
Given:
Initial velocity,

u = 0

(Since the ball dropped from rest)
Acceleration,

a = 10 m s^{- 2}

Distance (height)

s = 20 m

Final velocity

v = ?

Now,

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

Where

v, u, a,

and

s

are final velocity, initial velocity, acceleration and displacement, respectively.

{\Rightarrow v}^{2} = {(0)}^{2} + 2 \times 10 \times 20

\Rightarrow v = \sqrt{400}

\Rightarrow v = 20 m / s

Time:

v = u + at

Where

v, u, a,

and

t

are final velocity, initial velocity, acceleration and time, respectively.

\Rightarrow 20 = 0 + 10 t

\Rightarrow t = 2 s

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