A projectile is launched with initial speed v0 at angle θ. The horizontal range on level ground is given by which expression?

• A. R = v0 cos θ * t_f
• B. R = v0^2 sin θ / g
• C. R = v0^2 sin(2θ) / g
• D. R = (v0^2 cos^2 θ)/g

Answer: (C)

The main idea is that horizontal range comes from how far you travel horizontally while you’re in the air. The horizontal speed is v0 cos θ, and the total time in the air is the flight time t_f. So the range is R = v0 cos θ × t_f. On level ground, the flight time comes from the vertical motion: y(t) = v0 sin θ t − (1/2) g t^2. Setting y = 0 (excluding t = 0) gives t_f = 2 v0 sin θ / g. Substituting this into R = v0 cos θ t_f yields R = (v0 cos θ)(2 v0 sin θ / g) = v0^2 sin(2θ) / g, since sin 2θ = 2 sin θ cos θ.

That explicit form shows why the range depends on the square of the speed, the gravity, and the angle through sin(2θ). The expression that uses t_f is incomplete by itself because it doesn’t express R purely in terms of v0, θ, and g. The other options don’t reproduce the correct angular dependence: they either miss the sin 2θ factor or give nonphysical behavior (for example, predicting nonzero range at angles where the projectile barely leaves the ground).