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 when and where the two stones will meet.
Answers
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.
_/\_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.☺
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