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A boy standing on a cliff of $480 m$ height, drops a stone. One second later, he throws a second stone after the first. They both hit the ground at the same time. With what speed does he throw the second stone? (Take $= 10 m / s^{2}$ )

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5 m / s

10.5 m / s

14.7 m / s

19.6 m / s

answer is B.

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Detailed Solution

He throw the second stone with the speed of

10.5 m / s

.
Given data and assumptions:
Dropping height,

h = 480 m

Initial velocity of first stone,

u_{1} = 0

(as it is dropped freely.)
Initial velocity of 2nd stone,

= u_{2}

Using 2nd equation of motion for the first stone

h = u_{1} t + \frac{1}{2} g t^{2}

480 = 0 \times t + \frac{1}{2} \times 10 \times t^{2}

t^{2} = \frac{480}{5}

t^{2} = 96

t = 9.8 s

The time taken by the second stone will be

1

second less than the first stone.

t = 8.8 s

Using

2

nd equation of motion for the

2

nd stone

h = u_{2} t + \frac{1}{2} g t^{2}

480 = u_{2} \times (8.8) + \frac{1}{2} \times (10) \times (8.8)^{2}

\Rightarrow

u = 10.5 m / s

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