a stone is fallen from top of a tower 100m .at the same time another stone is projected vertically upwards from the ground with a velocity of 25m/s.calculate when and where the 2 stones will meet.
Answers
here is your solution
given,
height = 100m
velocity = 25m/sec
Now......
Suppose the two stones meet at a height ‘x’ from the ground.
The stone that is dropped from above will cover (100-x) before meeting the stone that is thrown from ground.
For the stone that is thrown from above the time taken to cover (100-x) is say, t.
(100-x) = (0)(t) + ½ (10)t2
=> 100 – x = 5t2 …………….(1)
For the stone that is thrown from the ground will cover distance ‘x’ in the same time ‘t’.
So, x = 25t – ½ (10)t2
=> x = 25t -5t2 ………………(2)
(1) + (2) => 100 = 25t
=> t = 4 s
Thus, the two stones will meet 4 s from the time of projection.
Using (2),
x = (25)(4) – (5)(42)
=> x = 100 – 80
=> x = 20 m
The stones meet 20 m above the ground.
I hope helps you
_/\_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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