A stone is allowed to fall from the top of a tower 100m high and at the same time another stone is projected veetically upwards from the ground with a velocity of 25m/s. Calculate when and where the stones will met?
(take g = 10m/s^2)
Answers
= 100 / (25 - 0)
= 4 seconds
Both will meet after 4s from start.
Distance travelled by projected stone after 4s
S = ut + 0.5at^2
S = 25*4 + 0.5*(-10)*4^2
S = 20 m
Stones will meet 20 m above the ground.
_/\_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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