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=\frac{u-v+\sqrt{u^{2}+v^{2}}}{g} $
B) $ t=\frac{u-v+\sqrt{u^{2}-v^{2}}}{g} $
C) $ t=\frac{u+v+\sqrt{u^{2}-v^{2}}}{g} $
D) $ t=\frac{u-v+\sqrt{u^{2}-v^{2}}}{2g} $
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 )-\frac{1}{2}g{{( t+T )}^{2}} $
And the distance moved by the second body in time $ T=vT-\frac{1}{2}gT^{2}=h $ (supposed above) ?(1)
$ \therefore $ The two bodies meet each other.
$ \therefore $ They are equid it’stant from the point of projection.
Hence, $ u( t+T )-\frac{1}{2}g{{( t+T )}^{2}}=vT-\frac{1}{2}gT^{2} $ … (2)
Also from (1) we get, $ h=vT-\frac{1}{2}gT^{2} $
$ \therefore \frac{dh}{dT}=v-gT $
$ \therefore $ h increases as T increases
$ \therefore $ T is minimum when h is minimum i.e., when $ \frac{dh}{dt}=0,i.e.,\text{ when }v-gT=0\text{ or }T=v/g. $
Substituting this value of T in (2), we get $ gt^{2}+2t( v-u )+2( v-u )( v/g )=0 $
or $ t=\frac{2g( v-u )+\sqrt{4g^{2}( v-u )+8vg^{2}( v-u )}}{2g^{2}} $
or $ t=\frac{u-v+\sqrt{u^{2}-v^{2}}}{g} $
Neglecting the negative sign which gives negative value of t.
