The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projecti

Question

The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projectile is:

Accepted Answer

1. **Formulas:**
   - Maximum Height ($H_m$): $H_m = \frac{v_0^2 \sin^2 	heta}{2g}$ .
   - Horizontal Range ($R$): $R = \frac{v_0^2 \sin 2	heta}{g}$ .
2. **Condition:** The problem states that the horizontal range equals the maximum height ($R = H_m$).
3. **Calculation:**
   $$ \frac{v_0^2 \sin 2	heta}{g} = \frac{v_0^2 \sin^2 	heta}{2g} $$
   Using the identity $\sin 2	heta = 2 \sin 	heta \cos 	heta$:
   $$ 2 \sin 	heta \cos 	heta = \frac{\sin^2 	heta}{2} $$
   Canceling terms ($v_0^2, g, \sin 	heta$) assuming $	heta 
eq 0$:
   $$ 2 \cos 	heta = \frac{\sin 	heta}{2} $$
   $$ 4 = \frac{\sin 	heta}{\cos 	heta} = 	an 	heta $$
   $$ 	heta = 	an^{-1}(4) $$

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