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➡A ball is dropped from the top of a tower 100m high and at the same Tim another ball is projected vertically upwards from the ground with a velocity of 25m/s. Find the height where the two balls will meet?

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Answer 1

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⬇⬇HERE IS UR ANSWER⬇⬇

They will meet instinctively. Here's how.

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.✔✔✅✅

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Answer 2

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.