Projectile Motion Analyzed — Question 4

Question 4

A fixed speed can give two launch angles for one target

A target lies 30 m away on the launch-height plane. Launch speed is fixed at 20 m/s, with an upward angle between $0^\circ$ and $90^\circ$. Neglect air resistance and use uniform $g=10\ \mathrm{m/s^2}$ downward. Unless stated otherwise, x is horizontal and y upward.

1. Find both launch angles that reach the target.

2. Find the flight time for each solution.

3. Can the same speed reach a level target 45 m away? Give the maximum-range argument.

Question 4 - Solution

A fixed speed can give two launch angles for one target

(a) Solve the sine equation with both branches.

$$\sin2\theta=\frac{gR}{v_0^2}=\frac{300}{400}=.75\ \Rightarrow\ \boxed{\theta_1=24.295^\circ,\quad\theta_2=65.705^\circ}$$ $30=(20^2/10)\sin2\theta$ gives $\sin2\theta=0.75$. Therefore $\theta_1=\tfrac12\sin^{-1}(0.75)\approx24.295^\circ$ and $\theta_2=90^\circ-\theta_1\approx65.705^\circ$. Taking only the principal inverse-sine result would miss the steep trajectory.

(b) Evaluate time separately for each angle.

$$T_i=\frac{R}{v_0\cos\theta_i}:\quad T_1=1.646\,\mathrm s,\quad T_2=3.646\,\mathrm s$$ $T=30/(20\cos\theta)$ gives approximately 1.646 s for the lower angle and 3.646 s for the higher angle. Substitution into $y=v_0\sin\theta\,T-5T^2$ returns zero for each.

(c) Apply the maximum possible sine.

$$R_{\max}=\frac{v_0^2}{g}=40\,\mathrm m<45\,\mathrm m$$ $R\leq v_0^2/g=40$ m, attained at $45^\circ$. A 45 m target would require $\sin2\theta=1.125$, impossible for a real angle. More launch speed, a different landing height or different physical assumptions would be required.