Question

A stone is dropped from the top of tower 100m high. Simultaneously another stone is projected upwards from ground a Velocity of 25 m/s. Find when and where two stones will meet. (Take g=10)

Answer 1

let the distance travelled by stone falling down be s
let the distance travelled by stone thrown up be 100-s
let the time when they meet be t
(1) For stone dropped from Tower
s= ut+1/2gt^2
s = 0×t + 1/2×(10m/s^2)×t^2
s= 5t^2 (i)
(2) for stone projected upwards
s= ut +1/2gt^2
100-s = 25×t+1/2 (-10)×t^2
100-s=25t-5t^2 (ii)
add (i)and (ii)
100-s+s=25t-5t^2+5t^2
100=25t
t=4 secs
constitute value of t in equation (i)
s= 5 (4)^2
s=5×16
s= 80m
hence both stone meet at the height of 80 m after 4 secs
HOPE IT HELPS! ! ! !

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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