Question

for which two angles of projection, can a body have the same horizontal range?
a 0 and 45°
b 0 and 45°
c none of the above
d 0 and 90°

Explanation:

Step1: Recall range formula

The horizontal range formula for projectile motion is $R=\frac{u^{2}\sin2\theta}{g}$, where $u$ is the initial velocity, $\theta$ is the angle of projection and $g$ is the acceleration due to gravity.

Step2: Analyze when ranges are equal

If we have two angles $\theta_1$ and $\theta_2$, and $R_1 = R_2$, then $\frac{u^{2}\sin2\theta_1}{g}=\frac{u^{2}\sin2\theta_2}{g}$. This implies $\sin2\theta_1=\sin2\theta_2$. We know that $\sin\alpha=\sin(180^{\circ}-\alpha)$. So, if $2\theta_1 = 2\theta_2$ or $2\theta_1=180^{\circ}- 2\theta_2$. The non - trivial case is when $\theta_2 = 90^{\circ}-\theta_1$. When $\theta_1=\theta$ and $\theta_2 = 90^{\circ}-\theta$, the ranges are equal. For example, if $\theta = 30^{\circ}$, then $90^{\circ}-\theta=60^{\circ}$, and a body projected at $30^{\circ}$ and $60^{\circ}$ with the same initial velocity will have the same horizontal range. Also, when $\theta = 0^{\circ}$, $\sin(2\times0^{\circ}) = 0$ and when $\theta = 90^{\circ}$, $\sin(2\times90^{\circ})=\sin180^{\circ}=0$. So, a body projected at $0^{\circ}$ and $90^{\circ}$ has zero range (same range in this special case of zero range).

Answer:

D. $0^{\circ}$ and $90^{\circ}$