For a projectile projected at angles $(45^{\circ}- heta)$ and $(45^{\circ}+ heta)$, the horizontal ranges de

Question

For a projectile projected at angles $(45^{\circ}-	heta)$ and $(45^{\circ}+	heta)$, the horizontal ranges described by the projectile are in the ratio of:

Accepted Answer

1. **Formula for Horizontal Range:** The horizontal range $R$ of a projectile projected with initial velocity $v_0$ at an angle $	heta_0$ is given by $R = \frac{v_0^2 \sin 2	heta_0}{g}$ .
2. **Analyze First Angle:** Let $	heta_1 = 45^{\circ} - 	heta$.
   $$ R_1 = \frac{v_0^2 \sin 2(45^{\circ} - 	heta)}{g} = \frac{v_0^2 \sin (90^{\circ} - 2	heta)}{g} $$
   Using the identity $\sin(90^{\circ} - x) = \cos x$, we get $R_1 = \frac{v_0^2 \cos 2	heta}{g}$.
3. **Analyze Second Angle:** Let $	heta_2 = 45^{\circ} + 	heta$.
   $$ R_2 = \frac{v_0^2 \sin 2(45^{\circ} + 	heta)}{g} = \frac{v_0^2 \sin (90^{\circ} + 2	heta)}{g} $$
   Using the identity $\sin(90^{\circ} + x) = \cos x$, we get $R_2 = \frac{v_0^2 \cos 2	heta}{g}$.
4. **Conclusion:** Since $R_1 = R_2$, the ranges are equal. The ratio is 1:1. This confirms the principle that for elevations which exceed or fall short of $45^{\circ}$ by equal amounts, the ranges are equal .

Step-by-Step Solution

Practice All PHYSICS Questions