Question

A stone is released from the top of a tower 90 m high at the same instant a second stone is projected vertically upwards from the ground with a velocity of 30 m/s when and where will the two stones meet.
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Answer 1

Answer:

T=3sec

Height=45m

__________

Answer 2

S1 = x

Here, u1 = 0

S1 = u1t + ½ at²

x = ½ at²

x = ½ ( 9.8 ) t²

x = 4.9t² ........(I)

Other stone is projected vertically upwards with velocity u2

Now, S2 = 90 - x

S2 = u2t + ½ at²

90 - x = ½ (9.8)t²

90 - x = 30t - 4.9t² ......(ii)

Adding (I) and (ii), we get,

x + 90 - x = 30t - 4.9t² + 4.9t²

90 = 30t

t = 3 seconds

Substituting t = 3 in (I), we get,

x = 4.9 x 9

= 44.1 m ( from the top )