Question

A stone is allowed to fall down 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 25 ms-1. Calculate when and where the two stones will meet?(take g = 10ms-2) how to do it ???i didnt understand.....

Answer 1

let "t" = time after which both stones meet
"S" = distance travelled by the stone dropped from the top of tower
(100-S) = distance travelled by the projected stone.

◆ i) For stone dropped from the top of tower
-S = 0 + 1/2 (-10) t²
or, S = 5t²

◆ ii) For stone projected upward
(100 - S) = 25t + 1/2 (-10) t²
= 25t - 5t²

Adding i) and ii) , We get
100 = 25t
or t = 4 s

Therefore, Two stones will meet after 4 s.

¡¡¡) Put value of t = 4 s in Equation i) , we get 

S = 5 × 16 

= 80 m.


Thus , both the stone will meet at a distance of 80 m from the top of tower.

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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