from top of a tower 100 m height a ball is dropped and at same instant another ball is projected vertically upwards from ground with velocity 25 m/s , find when and where two balls will meet.
Answers
Lets take distance covered by the stone thrown upwards to meet the other as x. Then the other stone covers a distance of (100-x)m.
Let the ball dropped down be b1. So the ball thrown up is b2.
b1 => d = 100-x
g = 9.8 m/s sq.
u = 0
We know, s= ut + 1/2at sq.
Putting values,
100-x = 4.9t sq. ....(1)
b2 => d = x
g = -9.8 m/s sq.
u = 25 m/s
We know, s= ut + 1/2at sq.
Putting values,
x = 25t -4.9t sq. ....(2)
Adding (1) and (2),
100-x +x = 4.9t sq. + 25t - 4.9t sq.
t = 4 secs.
Puuting t = 4 in (1),
100-x = 4.9t sq.
100-x =19.6
x = 80.4 m
So, they meet at 80.4 m from ground after 4 seconds
_/\_Hello mate__here is your answer--
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⚫Let the two bolls meet after a time t.
CASE 1 :-When the bolls 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 bolls thrown upwards
u = 25 ms−1
g = −9.8 ms−2(upward direction)
Let the displacement of the bolls 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 bolls has covered a distance given by (1) as = 4.9 × 4^2 = 78.4 m
Therefore, the bolls 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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