Physics, asked by rohit112, 1 year ago

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

Answered by Deekshii1
5
heya




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


rohit112: no this answer was wrong
rohit112: correct answer is - 78.4m from top, 4s
Answered by Anonymous
5

_/\_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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