Kinematics Question 618

Question: A body is projected vertically upwards with a velocity u, after time t another body is projected vertically upwards from the same point with a velocity v, where v < u. If they meet as soon as possible, then choose the correct option

Options:

A) t=uv+u2+v2g

B) t=uv+u2v2g

C) t=u+v+u2v2g

D) t=uv+u2v22g

Show Answer

Answer:

Correct Answer: B

Solution:

[b] Let the two bodies meet each other at a height h after time T of the projection of second body .

Then before meeting, the first body was in motion for time (t + T) whereas the second body was in motion for time T.

The distance moved by the first body in time (t+T)

=u(t+T)12g(t+T)2

And the distance moved by the second body in time T=vT12gT2=h (supposed above) ?(1)

The two bodies meet each other.

They are equid it’stant from the point of projection.

Hence, u(t+T)12g(t+T)2=vT12gT2 … (2)

Also from (1) we get, h=vT12gT2

dhdT=vgT

h increases as T increases

T is minimum when h is minimum i.e., when dhdt=0,i.e., when vgT=0 or T=v/g.

Substituting this value of T in (2), we get gt2+2t(vu)+2(vu)(v/g)=0

or t=2g(vu)+4g2(vu)+8vg2(vu)2g2

or t=uv+u2v2g

Neglecting the negative sign which gives negative value of t.



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