A mortar fires shells at a target B with initial speed v_0 =…
A mortar fires shells at a target B with initial speed $v_0 = 150\ \mathrm{m/s}$. The target is hit both with the launch angle $\alpha = 30^\circ$ and with $\beta = 60^\circ$. How far away is the target (level ground, no air resistance)?
Explanation of Method
Two launch angles that hit the same target.
Step 1: Range formula
\[ R = \frac{v_0^2\sin 2\alpha}{g}. \]
Step 2: Both angles
\[ R_{30} = \frac{22\,500 \cdot \sin 60^\circ}{9.81}, \]
\[ R_{60} = \frac{22\,500 \cdot \sin 120^\circ}{9.81}, \]
\[ \sin 60^\circ = \sin 120^\circ = 0.866. \]
\[R \approx 1986\ \mathrm{m} \ \text{(for both angles)}\]
How the two shots differ
| Angle | Flight time $\tfrac{2v_0\sin\alpha}{g}$ | Highest point $\tfrac{v_0^2\sin^2\alpha}{2g}$ |
|---|---|---|
| 30° | 15.3 s | 287 m |
| 60° | 26.5 s | 860 m |
- fired at 30°
- fired at 60°
Result
\[\boxed{\,R \approx 1986\ \mathrm{m} \ \text{(for both angles)}\,}\]
Remark
Check. The original result is correct: complementary angles ($\alpha + \beta = 90^\circ$) always give the same range, because $\sin 2\alpha = \sin(180^\circ - 2\alpha)$. The label "B12" in the statement probably means the single target B hit by shells 1 and 2.