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A projectile is fired vertically upwards with an initial velocity $u$. After an interval of $T$ seconds a second projectile is fired vertically upwards, also with initial velocity $u$.
They meet at time $t = \frac{u}{g}$ and at a height $\frac{{{u^2}}}{{2g}} + \frac{{g{T^2}}}{8}$
They meet at time $t = \frac{u}{g} + \frac{T}{2}$ and at a height $\frac{{{u^2}}}{{2g}} + \frac{{g{T^2}}}{8}$
They meet at time $t = \frac{u}{g} + \frac{T}{2}$ and at a height $\frac{{{u^2}}}{{2g}} - \frac{{g{T^2}}}{8}$
They never meet
Solution
(c) For first projectile, ${h_1} = ut – \frac{1}{2}g{t^2}$
For second projectile, ${h_2} = u(t – T) – \frac{1}{2}g{(t – T)^2}$
When both meet i.e. ${h_1} = {h_2}$
$ut – \frac{1}{2}g{t^2} = u(t – T) – \frac{1}{2}g{(t – T)^2}$
==> $uT + \frac{1}{2}g{T^2} = gtT$
==> $t = \frac{u}{g} + \frac{T}{2}$
and ${h_1} = u\left( {\frac{u}{g} + \frac{T}{2}} \right) – \frac{1}{2}g{\left( {\frac{u}{2} + \frac{T}{2}} \right)^2}$
$ = \frac{{{u^2}}}{{2g}} – \frac{{g{T^2}}}{8}$.
