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A stone is allowed to fall from the top of a tower $100 m$ high and at the same time another stone is projected vertically upwards from the ground with a velocity of $25 m / s$ . calculate where and when the two stones will meet.

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100 m, 10 s

80 m, 5 s

80 m, 4 s

100 m, 5 s

answer is C.

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Detailed Solution

A stone is allowed to fall from the top of a tower

100 m

high and at the same time another stone is projected vertically upwards from the ground with a velocity of

25 m / s

. The two stones will meet after

4 s

at a height of

80 m

from the ground.
Given data in the problem,
Height of tower,

s = 100 m

For first stone Initial velocity,

u = 0

By using the second equation,

s = ut + \frac{1}{2} {at}^{2}

= > a = g

Where,

s = distance

u = initial velocity

a = uniform acceleration

t = time

By putting all the values in the above equation,

= > x = 0 + \frac{1}{2} g t^{2}

, where

x

is the height where two stones meet.

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

= > x = 5 {\times t}^{2}

… (1)
Now, for second stone Initial velocity,

u = 25 m / s

By using the second equation,

s = ut + \frac{1}{2} {at}^{2}

Where acceleration

a = g

(acceleration due to gravity) Where,

s = distance

u = initial velocity

a = uniform acceleration

t = time

By putting all the values in the above equation,

\Rightarrow 100 – x = ut + \frac{1}{2} g t^{2}

\Rightarrow 100 – x = (25) \times t + \frac{1}{2} \times 10 {\times t}^{2}

\Rightarrow 100 – x = 25 \times t - 5 {\times t}^{2}

… (2)
Now adding equation 1 and 2

\Rightarrow 25 t = 100

\Rightarrow t = 4 s

putting value of t in equation 1, we get

\Rightarrow x = 5 \times 42

\Rightarrow x = 80 m

The stones will meet up at a height of

80 m

and time is

4 s

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