Question

46. 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
1)T/2(2v - gT/v -gT)
2) T/3(3v - gT/ 2v - gT)
3) T/2( v - gT/2v- gT)
4)T/3(2v - gT/3v- gT)​

Answer 1

The time at which the two bodies will meet is T/2 (2v-gT/v-gT), option 1.

Given,

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

Initial velocity of body(2) =v

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

To find,

The time at which the two bodies meet =?

Solution,

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

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

Let the height of the tower is=h

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

According to law of kinematics:

s= ut + 1/2 gT²

x= 0+ 1/2gT²

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

x= vT+ 1/2gt²

Solving both equation we get T/2(2v- gT/v - gt)

Hence, the time at which the two bodies will meet is T/2(2v- gT/v-gT).

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