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

We know that the flight time of the first projectile is $t=\frac{v_y1}{g}=\frac{vsin(45+\alpha)}{g}$ The flight time of the second projectile is given as $t_a=\frac{v_y2}{g}=\frac{vsin(45-\alpha)}{g}$ Now we can find the ratio as $\frac{t_1}{t_2}=\frac{\frac{vsin(45+\alpha)}{g}}{\frac{vsin(45-\alpha)}{g}}=\frac{vsin(45+\alpha)}{vsin(45-\alpha)}=tan(45+\alpha)$