Projectile Motion Analyzed — Question 7

Question 7

Minimize launch speed by choosing the flight time

A point target is at $(12,5)$ m relative to launch. You may choose any positive arrival time T and launch direction. Neglect air resistance and use uniform $g=10\ \mathrm{m/s^2}$ downward. Unless stated otherwise, x is horizontal and y upward. For this task minimize the initial speed; there are no obstacles.

1. Express the required launch velocity components and squared speed as functions of T.

2. Find the minimum speed and the corresponding arrival time, using calculus or $a+b\geq2\sqrt{ab}$ for positive a,b.

3. Find the launch angle and explain why the minimum is unique in positive T.

Question 7 - Solution

Minimize launch speed by choosing the flight time

(a) Solve the endpoint constraints for the launch components.

$$v_{0x}=12/T,\quad v_{0y}=5/T+5T,\quad v_0^2=169/T^2+50+25T^2\quad\text{(SI numerical }T)$$ $v_{0x}=12\ \mathrm m/T$ and $v_{0y}=5\ \mathrm m/T+(5\ \mathrm{m/s^2})T$. With $\tau=T/(1\ \mathrm s)$, $v_0^2=[169/\tau^2+50+25\tau^2]$ m$^2$/s$^2$.

(b) Balance the two positive variable terms.

$$\frac{169}{T^2}+25T^2\ge2\sqrt{169(25)}=130\ \Rightarrow\ \boxed{T=\sqrt{13/5}\,\mathrm s,\quad v_{0,\min}=6\sqrt5\,\mathrm{m/s}}$$ $169/\tau^2+25\tau^2\geq2\sqrt{169\cdot25}=130$. Equality requires $\tau^2=13/5$. Thus $T=\sqrt{13/5}\approx1.612$ s and $v_{0,\min}=\sqrt{180}=6\sqrt5\approx13.416$ m/s.

(c) Recover the optimum direction.

$$\tan\theta=\frac{5+5T^2}{12}=\frac32\ \Rightarrow\ \theta=56.310^\circ,\quad169/T^2=25T^2\text{ fixes }T>0$$ $\tan\theta=v_{0y}/v_{0x}=(5+5\tau^2)/12=18/12=3/2$, so $\theta\approx56.310^\circ$. The equality condition fixes one positive T; at either T approaching zero or infinity the required speed diverges. The minimum therefore selects a unique time and launch vector in this model.