Kinematics: Tricky Questions Solved (5)

Let’s pick particle 1 (the projectile) as the observer. The dropped particle falls only under gravity. Both have $g$ downward, so in the frame of particle 1, the dropped particle has zero relative acceleration.

Initial velocity of projectile: $u_x = 20\cos 60^\circ = 10$ m/s, $u_y = 20\sin 60^\circ = 10\sqrt{3}$ m/s. The dropped particle starts at rest. Relative velocity of dropped particle w.r.t. projectile is $(-10, -10\sqrt{3})$ m/s.

The dropped particle starts $40$ m vertically above. Its relative position at time $t$:

We need $|\Delta x| = |\Delta y|$:

For small $t$, the right side is positive: $10 t = 40 - 10\sqrt{3}\, t \Rightarrow t(10 + 10\sqrt{3}) = 40$.

Final answer: $t = 2(\sqrt{3}-1) \approx 1.46$ s.

Why This Works

Both particles share the same gravitational acceleration. In the relative frame, gravity vanishes and motion becomes a straight line at constant velocity. This is the single biggest time-saver in two-body kinematics.

You can attempt this problem with absolute coordinates too, but you would burn three minutes on algebra. The relative-velocity trick gets you there in under a minute.

Alternative Method

Using absolute coordinates: projectile at $(10t,\, 10\sqrt{3}t - 5t^2)$, dropped particle at $(0,\, 40 - 5t^2)$. Set horizontal separation equal to vertical separation. The $-5t^2$ terms cancel and you reach the same equation.