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

Solution :
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 $$\left( {t + T} \right)$$  whereas the second body was in motion for time $$T.$$
The distance moved by the first body in time $$\left( {t + T} \right)$$
$$ = u\left( {t + T} \right) - \frac{1}{2}g{\left( {t + T} \right)^2}.$$
And the distance moved by the second body in time
$$T = vT - \frac{1}{2}g{T^2} = h\,\,\left( {{\text{supposed above}}} \right).\,......\left( 1 \right)$$
$$\because $$ The two bodies meet each other,
$$\therefore $$ They are equidistant from the point of projection.
$$\eqalign{ & {\text{Hence,}}\,\,\,u\left( {t + T} \right) - \frac{1}{2}g{\left( {t + T} \right)^2} = vT - \frac{1}{2}g{T^2} \cr & {\text{or}}\,\,g{t^2} + 2t\left( {gT - u} \right) + 2\left( {v - u} \right)T = 0\,......\left( 2 \right) \cr} $$
Also from (1) we get,
$$h = vT - \frac{1}{2}g{T^2}$$
$$\therefore \frac{{dh}}{{dT}} = v - gT$$
$$\therefore $$ $$h$$ increases as $$T$$ increases
$$\therefore $$ $$T$$ is minimum when $$h$$ is minimum i.e., when
$$\frac{{dh}}{{dT}} = 0,\,\,{\text{i}}{\text{.e}}{\text{.}}\,{\text{when}}\,\,v - gT = 0\,\,{\text{or}}\,\,T = \frac{v}{g}.$$
Substituting this value of $$T$$ in (2), we get
$$\eqalign{ & g{t^2} + 2t\left( {v - u} \right) + 2\left( {v - u} \right)\left( {\frac{v}{g}} \right) = 0 \cr & {\text{or}}\,\,t = \frac{{2g\left( {u - v} \right) + \sqrt {4{g^2}{{\left( {u - v} \right)}^2} + 8v{g^2}\left( {u - v} \right)} }}{{2{g^2}}} \cr & {\text{or}}\,\,t = \frac{{u - v + \sqrt {{u^2} - {v^2}} }}{g} \cr} $$
neglecting the negative sign which gives negative value of $$t.$$