A projectile is fired from a cannon with speed v_0 = 400 m/s…
A projectile is fired from a cannon with speed $v_0 = 400\ \mathrm{m/s}$ at $\alpha = 30^\circ$ to the horizontal. Find its horizontal range and its time of flight (level ground, no air resistance).
Explanation of Method
Range and flight time of a cannon shell.
Step 1: Components
\[ v_{0x} = 400\cos 30^\circ = 200\sqrt3 \approx 346.4\ \mathrm{m/s}, \]
\[ v_{0y} = 400\sin 30^\circ = 200\ \mathrm{m/s}. \]
Step 2: Time of flight
It lands when $y = v_{0y}t - \tfrac12 g t^2 = 0$ again: $T = \frac{2v_{0y}}{g} = \frac{400}{9.81}$.
\[T \approx 40.8\ \mathrm{s}\]
Step 3: Range
\[ R = v_{0x}T = \frac{v_0^2\sin 2\alpha}{g} = \frac{160\,000 \cdot 0.8660}{9.81}. \]
\[R \approx 14\,125\ \mathrm{m} \approx 14.1\ \mathrm{km}\]
Result
\[\boxed{\,T \approx 40.8\ \mathrm{s}\,}\]
\[\boxed{\,R \approx 14\,125\ \mathrm{m} \approx 14.1\ \mathrm{km}\,}\]
Remark
Note. The original 14 128.44 m is a small arithmetic slip ($138\,564/9.81 = 14\,125$). The highest point is $\tfrac{v_{0y}^2}{2g} \approx 2039\ \mathrm{m}$, reached after 20.4 s. In reality air resistance would reduce the range a lot at such speeds.