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 arey(t)=votsin(θ)−12gt2andx(t)=votcos(θ)Setting y(t)=0 yields 0=votsin(θ)−12gt2. If we solve for t, we obtain, by factoring,t=2vsin(θ)gSubstitute this into our equation for x(t). This yieldsx(t)=2v2cos(θ)sin(θ)gThis is equal to x=v^2sin(2θ)gHence the angles that have identical projectiles are have the same range via substitution in the last equation is C.  60.23°, 29.77° 

It may possibly be c i think |-/