Question

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

Answer 1

Consider, a stone fall from the top of a tower (h) and distance covered by a stone at time  t

Height (h) = 100 m

So,  x - x0 = u0t + 1/2 gt2

Initial velocity  = 0

or, 100 - x = 0 +1/2 gt2 .....eq (1)

Distance covered by stone from ground

Where initial velocity (u0) = 25 m/s

x = u0t  - 1/2 gt2

or, x =( 25 X t) - 1/2 gt2  .....eq (2)

Combine .....eq (1) and .....eq (2)

So, 100 = 25t

t = 4

therefore, x = 25X4 -1/2 X9.8X 42 = 100 - 78.4 = 21.6 m

Answer 2

_/\_Hello mate__here is your answer--

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⚫Let the two stones meet after a time t.

CASE 1 :-When the stone dropped from the tower

u = 0 m/s

g = 9.8 ms−2

Let the displacement of the stone in time t from the top of the tower be s.

From the equation of motion,

s = ut + 1/2gt^2

⇒s = 0 × + 1/2× 9.8 ×t ^2

⇒ s = 4.9t^2 …………………… . (1)

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CASE 2 :--When the stone thrown upwards

u = 25 ms−1

g = −9.8 ms−2(upward direction)

Let the displacement of the stone from the ground in time t be '

Equation of motion,

s' = ut+ 1/2gt^2

⇒s′ = 25 × − 1/2× 9.8 × t^2

⇒s′ = 25 − 4.9t^2 …………………… . (2)

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Given that the total displacement is 100 m.

s′ + s = 100

⇒ 25 − 4.9t^2 + 4.9t^2 = 100

⇒ t =100 /25 = 4 s

The falling stone has covered a distance given by (1) as = 4.9 × 4^2 = 78.4 m

Therefore, the stones will meet after 4 s at a height (100 – 78.4) = 20.6 m from the ground.

I hope, this will help you.☺

Thank you______❤

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