A projectile is launched with speed v0 at angle θ above the horizontal. On level ground, what is the expression for its maximum horizontal range R?

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

Answer: (C)

When a projectile is launched from and lands at the same height on level ground, its horizontal range comes from how long it stays in the air and how fast it moves horizontally. The vertical motion determines the time of flight: y(t) = v0 sinθ t − (1/2) g t^2. Setting y = 0 (after launch) gives the nonzero time of flight t = 2 v0 sinθ / g. The horizontal range is the horizontal speed times this time: R = v0 cosθ × (2 v0 sinθ / g) = (v0^2 / g) sin 2θ. Thus the range is R = v0^2 sin 2θ / g, which is also equal to 2 v0^2 sinθ cosθ / g by the sine double-angle identity. The maximum range occurs when sin 2θ is largest, at θ = 45°, giving R_max = v0^2 / g.