A cannon with a muzzle speed of 1000 m/s is fired at a target 2000 m away horizontally and 800 m above the cannon. At what angle above the horizontal should the cannon be fired (neglect air resistance)?

What you are solving

The target is not at the same height as the cannon, so you need the projectile equation that includes a vertical displacement $y$ at a given horizontal range $x$.

Write the trajectory equation

For a launch speed $v_0$ at angle $\theta$ (no air resistance):

• Horizontal: $x=v_0\cos\theta\, t$
• Vertical: $y=v_0\sin\theta\, t-\dfrac{1}{2}gt^2$

Eliminate $t$ using $t=\dfrac{x}{v_0\cos\theta}$:

$$ y=x\tan\theta-\frac{g x^2}{2v_0^2\cos^2\theta} $$

Convert to a quadratic in $\tan\theta$

Let $T=\tan\theta$ and use $\dfrac{1}{\cos^2\theta}=1+\tan^2\theta=1+T^2$. Define $$ A=\frac{g x^2}{2v_0^2} $$ Then $$ y=xT-A(1+T^2) $$ Rearrange: $$ AT^2-xT+(y+A)=0 $$

Substitute numbers and solve

Given $v_0=1000$, $x=2000$, $y=800$, $g=9.8$:

$$ A=\frac{9.8\,(2000)^2}{2(1000)^2}=19.6 $$ So $$ 19.6T^2-2000T+819.6=0 $$ Using the quadratic formula, $$ T=\frac{2000\pm\sqrt{2000^2-4(19.6)(819.6)}}{2(19.6)}\approx\frac{2000\pm 1983.9}{39.2} $$ This gives $$ T_1\approx 0.4115,\quad T_2\approx 101.6 $$ Convert back to angles: $$ \theta_1=\arctan(0.4115)\approx 22.4^\circ,\quad \theta_2=\arctan(101.6)\approx 89.4^\circ $$

Interpreting the two angles

There are typically two launch angles that hit the same point: a lower, faster, flatter shot ($\approx 22.4^\circ$) and a very high, lobbed shot ($\approx 89.4^\circ$).