Question

A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet.









Please give me a short answer




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Answer 1

Let the stones meet at point A after time t.

For upper stone :

u

′

=0

x=0+ 21 gt 2

x= 21 ×10×t 2

⟹x=5t 2

(1)

For lower stone :

u=25 m/s

100−x=ut− 21 gt 2

100−x=(25)t− 21×10×t 2

⟹100−x=25t−5t

2

.(2)

Adding (1) and (2), we get

25t=100

⟹t=4 s

From (1),

x=5×4 2

⟹x=80 m

Hence the stone meet at a height of 20 m above the ground after 4 seconds.