Step-by-step explanation:
Projectiles launched at complementary angles have the same range. Only option D has complementary angles (adds up to 90°).
We can prove this with kinematics.
First, find the time for a projectile to land:
Δy = v₀ t + ½ at²
0 = (v₀ sin θ) t + ½ (-g) t²
0 = v₀ sin θ − ½ g t
t = 2 v₀ sin θ / g
Now find the horizontal distance (range) traveled in that time:
Δx = v₀ t + ½ at²
Δx = (v₀ cos θ) (2 v₀ sin θ / g) + 0
Δx = 2 v₀² sin θ cos θ / g
Δx = v₀² sin (2θ) / g
If we replace θ with 90°−θ:
Δx = v₀² sin (2(90°−θ)) / g
Δx = v₀² sin (180°−2θ) / g
Δx = v₀² sin (2θ) / g
We get the same range.