A projectile is fired vertically upwards with an initial velocity… - Vidyadip

MCQ

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 $u$.

A
They meet at time $t = \frac{u}{g}$ and at a height $\frac{{{u^2}}}{{2g}} + \frac{{g{T^2}}}{8}$
B
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 $\frac{{{u^2}}}{{2g}} - \frac{{g{T^2}}}{8}$
D
They never meet

✓

Answer

Correct option: C.

They meet at time $t = \frac{u}{g} + \frac{T}{2}$ and at a height $\frac{{{u^2}}}{{2g}} - \frac{{g{T^2}}}{8}$

c
(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}$.

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