Physics Two Dimensional Motion Question 166
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} $
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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 equidistant 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.
BETA
