A projectile is fired at an angle of $45^{\circ}$ with the horizontal. The elevation angle $\alpha$ of the pro

Question

A projectile is fired at an angle of $45^{\circ}$ with the horizontal. The elevation angle $\alpha$ of the projectile at its highest point as seen from the point of projection is:

Accepted Answer

1. **Geometry of the Path:** Let the projectile be fired with initial velocity $v_0$ at an angle $	heta = 45^\circ$. The highest point ($P$) of the trajectory has coordinates $(x, y) = (R/2, H)$, where $R$ is the horizontal range and $H$ is the maximum height.
2. **Angle of Elevation:** The angle of elevation $\alpha$ is the angle made by the position vector of point $P$ with the horizontal. From trigonometry:
   $$ 	an \alpha = \frac{y}{x} = \frac{H}{R/2} = \frac{2H}{R} $$
3. **Formulas:**
   - Maximum Height: $H = \frac{v_0^2 \sin^2 	heta}{2g}$ .
   - Horizontal Range: $R = \frac{v_0^2 \sin 2	heta}{g} = \frac{2v_0^2 \sin 	heta \cos 	heta}{g}$ .
4. **Substitution:**
   $$ 	an \alpha = \frac{2 \left( \frac{v_0^2 \sin^2 	heta}{2g} 
ight)}{\frac{2v_0^2 \sin 	heta \cos 	heta}{g}} = \frac{\sin^2 	heta}{2 \sin 	heta \cos 	heta} = \frac{1}{2} 	an 	heta $$
5. **Calculation:** Given $	heta = 45^\circ$:
   $$ 	an \alpha = \frac{1}{2} 	an 45^\circ = \frac{1}{2}(1) = \frac{1}{2} $$
   $$ \alpha = 	an^{-1}\left(\frac{1}{2}ight) $$

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