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Question

From the foot of a tower 90 m high, a stone is thrown up so as to reach the top of the tower with zero velocity. Two seconds later another stone is dropped from the top of the tower. Find when will two stones meet for the first time ?

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Solution

The correct option is **C** 3.1 s

First stone is thrown so as to reach the top of the tower, so its initial velocity is given by

v2−u2=2(−g)H (By third equation of motion)

⇒0−u2=−2gH

⇒u=√2gH

=√2×10×90=42.5 m/s

Let us take the time t=t0 when the two stones meet at a height H above the foot of the tower. The first stone is travelled a height H in the duration t0 and the second stone has fallen a distance (90−H) in time (t0−2).

Using second equation of motion

For first stone : H=42.5t0−12(10)t20

For second stone: 90−H=12(10)(t0−2)2

Adding above two equation, we get

22.5t0=70

Or t0=3.11 s

First stone is thrown so as to reach the top of the tower, so its initial velocity is given by

v2−u2=2(−g)H (By third equation of motion)

⇒0−u2=−2gH

⇒u=√2gH

=√2×10×90=42.5 m/s

Let us take the time t=t0 when the two stones meet at a height H above the foot of the tower. The first stone is travelled a height H in the duration t0 and the second stone has fallen a distance (90−H) in time (t0−2).

Using second equation of motion

For first stone : H=42.5t0−12(10)t20

For second stone: 90−H=12(10)(t0−2)2

Adding above two equation, we get

22.5t0=70

Or t0=3.11 s

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