Question

A ball is dropped from the top of a tower 100 m high and at the same time another ball is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate where and when the two stones will meet. [Take g= 9.8 m/s²]​

Answer 1

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Answer 2

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⏩ Let the ball 'A' be dropped from the tower.

Its initial velocity u1= 0 and acceleration a1= +g = + 10 m/s².

The ball 'B' is projected upwards with velocity u2 = 25 m/s and it's acceleration is a2 = -g= -10 m/s².

Let the two balls meet at point 'C' at distance 'y' below from 'A'.

Let the ball 'A' be dropped from the tower

Now,

For ball A,

y= u1t + 1/2 × a1t²

=> y= 1/2 × 10t²

=> y= 5t² .......→ (i)

For ball B,

(100-y) = u2t + 1/2 × a2t²

=> (100 -y) = 25 × t + 1/2 × (-10) × t²

=> 100-y= 25t - 5t² .......→ (ii)

On adding equation (i) and (ii), we have,

100 = 25t

=> t= 4s

On substituting the value of t in (i), we get,

y= 5 × 4²= 80 m.

[From the top of the tower or 20 m from the ground].

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