At what angle must a body be thrown so that (a) its maximum…

At what angle must a body be thrown so that (a) its maximum height equals its horizontal range; (b) its horizontal range equals the height it would reach if thrown vertically upward with the same speed?

Explanation of Method

Angles for special heights and ranges.

Step 1: (a) Height equals range

\[ \frac{v_0^2\sin^2\alpha}{2g} = \frac{2v_0^2\sin\alpha\cos\alpha}{g} \;\Longrightarrow\; \sin\alpha = 4\cos\alpha. \]

\[\tan\alpha = 4, \qquad \alpha \approx 76.0^\circ\]

Step 2: (b) Range equals the vertical height

Thrown straight up, the body rises $v_0^2/(2g)$. Setting the range equal to it: \[\frac{v_0^2\sin 2\alpha}{g} = \frac{v_0^2}{2g} \;\Longrightarrow\; \sin 2\alpha = \frac12 \;\Longrightarrow\; 2\alpha = 30^\circ \text{ or } 150^\circ.\]

\[\alpha = 15^\circ \quad \text{or} \quad \alpha = 75^\circ\]

$\text{(a) } H = R$ $H$ $R$ $\alpha \approx 76^\circ$ $\text{(b) } R = H_{\text{vertical}}$ $\tfrac{v_0^2}{2g}$ $R = \tfrac{v_0^2}{2g}$ $15^\circ$ $75^\circ$

Both panels to scale with the same launch speed. (a) At $\tan\alpha = 4$ the height equals the range. (b) Two angles, 15° and 75°, give a range equal to the height of a vertical throw (grey bar).

Result

\[\boxed{\,\tan\alpha = 4, \qquad \alpha \approx 76.0^\circ\,}\]

\[\boxed{\,\alpha = 15^\circ \quad \text{or} \quad \alpha = 75^\circ\,}\]

Remark

About the original solution. Part (a) is correct. In part (b) it gave only 15°, but $\sin 2\alpha = \tfrac12$ also holds for $2\alpha = 150^\circ$; complementary angles always give the same range, so 75° works too (a higher, slower throw).