## Essential University Physics: Volume 1 (3rd Edition)

We know that the range of the first projectile is given as: $R_1=\frac{v_{\circ}^2 sin(2\theta)}{g}$ $R_1=\frac{v_{\circ}^2 sin(2(45+\alpha))}{g}=\frac{v_{\circ}^2 sin(90+2\alpha)}{g}$ We can find the range of the second projectile as: $R_2=\frac{v_{\circ}^2 sin(2\theta)}{g}$ $R_2=\frac{v_{\circ}^2 sin(2(45-\alpha))}{g}=\frac{v_{\circ}^2 sin(90-2\alpha)}{g}$ As we know that $sin(90+\theta)=sin(90-\theta)$, the range of the projectile is the same.