For which pair of launch angles will two identical projectiles have equal ranges?. A. 19.24°, 80.54°B. 16.42°, 74.58°C. 60.23°, 29.77°D. 89.53°, 01.47°E. 42.42°, 47.59°

The kinematic equations of motion that apply here are y (t) = votsin (θ) - 12gt2 and x (t) = votcos (θ) Setting y (t) = 0 yields 0=votsin (θ) - 12gt2. If we solve for t, we obtain, by factoring, t = 2vsin (θ) g Substitute this into our equation for x (t). This yields x (t) = 2v2cos (θ) sin (θ) g This is equal to x = v^2sin (2θ) g Hence the angles that have identical projectiles are have the same range via substitution in the last equation is C. 60.23°, 29.77°