At time t = 0, a body is dropped freely from the top of a tall building and at a later time t = T, another body is thrown vertically downwards with a velocity ‘v’ from the top of the same building. The time at which the two bodies will meet is

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$\frac{T}{3} [\frac{2 v - gT}{3 v - gT}]$

$\frac{T}{3} [\frac{3 v - gT}{2 v - gT}]$

$\frac{T}{2} [\frac{2 v - gT}{v - gT}]$

$\frac{T}{2} [\frac{v - gT}{2 v - gT}]$

answer is A.

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Detailed Solution

Initial velocity of body(1) =0 (dropped freely)

Initial velocity of body(2) =v

Acceleration of the bodies =g =9.8m/s².

According to the question, the body is dropped freely at t=0.

∴ The initial velocity of the body 1 (u)=0.

Let the height of the tower is = h

Let distance traveled by body(1) in time T be x.

According to the law of kinematics:

s= ut + 1/2 gt²

x= 0+ 1/2gt²

Now, the time taken by the second body to cover the same distance x will be:

x= vT+ 1/2gt²

Solving both equation we get $\frac{T}{2} (\frac{2 v - g T}{v - g t})$

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