A projectile is projected from the origin with a speed of $20m{s^{ - 1}}$ at an angle of ${63^\circ }$ above the horizontal. After a delay of 1s, a drone is launched from the same point with constant velocity. The motion of both the projectile and the drone takes place in the vertical xz-plane. If the drone strikes the projectile 2s after the projectile is launched, find the velocity of the drone. Ignore air resistance. [$g = 10m{s^{ - 2}}$, $\sin {63^\circ } = \frac{4}{5}$ and $\cos {63^\circ } = \frac{3}{5}$]
Solution
For projectile motion,
$x = 20\cos {63^\circ }.2 = 24m$
$z = 20\sin {63^\circ }.2 - \frac{1}{2}{.10.2^2} = 32 - 20 = 12m$
Position vector of the projectile when the drone strikes it,
$\vec r = 24\hat i + 12\hat k$ m
Velocity of drone = $\frac{{\vec r}}{1} = 24\hat i + 12\hat k$ $m{s^{ - 1}}$