Q. 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 is
A) They meet at time $t = \frac{u}{g}$ and at a height $\frac{u^{2}}{2G} + \frac{gT^{2}}{8}$
B) They meet at time $t = \frac{u}{g} + \frac{T}{2}$ and at a height $\frac{u^{2}}{2g} + \frac{gT^{2}}{8}$
C) They meet at time $t = \frac{u}{g} + \frac{T}{2}$ and at a height $\frac{u^{2}}{2g} - \frac{gT^{2}}{8}$
D) They never meet
Explanation
(c) For first projectile, $h_{1} = ut - \frac{1}{2}gt^{2}$
For second projectile, $h_{2} = u(t - T) - \frac{1}{2}g(t - T)^{2}$
When both meet i.e. $h_{1} = h_{2}$
$ut - \frac{1}{2}gt^{2} = u(t - T) - \frac{1}{2}g(t - T)^{2}$
$\Rightarrow uT + \frac{1}{2}gT^{2} = gtT \Rightarrow t = \frac{u}{g} + \frac{T}{2}$
and $h_{1} = u\left( \frac{u}{g} + \frac{T}{2} \right) - \frac{1}{2}g{\ \left( \begin{array}{r}
\frac{u}{g} + \frac{T}{2} \\
\ \\
\
\end{array} \right)}^{2}$
$= \frac{u^{2}}{2g} - \frac{gT^{2}}{8}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.