Question

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°

Answer 1

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° 

Answer 2

The answer is C

If you subtract the given range of a projectile from 90º, It will give you its equivalent projectile counter part in terms of Range.

So, in this example where it gives you both the projectiles and asks you which ones that have the same range, add both and if it adds up to exaclt 90º, that is your answer

60.23º + 29.77º = 90º