Projectile Motion Analyzed — Question 3

Question 3

Complementary launches share a range but not a flight

Two projectiles are launched from level ground at 20 m/s, one at $30^\circ$ and the other at $60^\circ$. Both land at launch height. Neglect air resistance and use uniform $g=10\ \mathrm{m/s^2}$ downward. Unless stated otherwise, x is horizontal and y upward.

1. Derive their common range from the component equations.

2. Find each flight time and maximum height.

3. Which projectile reaches the common landing point first, and why does equal range not mean equal average velocity?

Question 3 - Solution

Complementary launches share a range but not a flight

(a) Combine the vertical return time with horizontal motion.

$$0=v_0\sin\theta\,T-\tfrac12gT^2\ \Rightarrow\ R=\frac{v_0^2}{g}\sin2\theta=\boxed{20\sqrt3\,\mathrm m}$$ For a nonhorizontal upward launch, $T=2v_0\sin\theta/g$ and $R=v_0\cos\theta\,T=(v_0^2/g)\sin2\theta$. Both angles give $R=40\sin60^\circ=20\sqrt3\approx34.641$ m.

(b) Retain each vertical launch component.

$$T_{30}=2,\quad T_{60}=2\sqrt3\quad\mathrm s;\qquad H_{30}=5,\quad H_{60}=15\quad\mathrm m$$ The $30^\circ$ flight has $T=2$ s and $H=5$ m. The $60^\circ$ flight has $T=2\sqrt3\approx3.464$ s and $H=15$ m. Their launch speeds agree, but their vertical shares differ.

(c) Divide the same displacement by different times.

$$\vec v_{{\mathrm{av}},30}=(R/2,0)=(10\sqrt3,0),\quad\vec v_{{\mathrm{av}},60}=(R/(2\sqrt3),0)=(10,0)\quad\mathrm{m/s}$$ The lower-angle launch arrives first. Its average velocity is $(10\sqrt3,0)$ m/s, whereas the higher-angle average is $(10,0)$ m/s. They share a horizontal endpoint displacement but not an elapsed time. This complementary-angle result assumes equal release and landing heights.