A projectile is launched from ground level with speed v0 at angle θ above the horizontal, neglecting air resistance. Which expression gives its horizontal range R in terms of v0, θ, and g?

A projectile is launched from ground level with speed v0 at angle θ above the horizontal, neglecting air resistance. Which expression gives its horizontal range R in terms of v0, θ, and g?

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

Answer: (B)

The key idea is that the horizontal range comes from multiplying the horizontal speed by the time the projectile is in the air. The horizontal speed is v0 cos θ, and for launch from ground level to land at the same level, the total flight time is T = 2 v0 sin θ / g. So the range is R = (v0 cos θ) × (2 v0 sin θ / g) = (2 v0^2 sin θ cos θ)/g. Using the identity sin 2θ = 2 sin θ cos θ, this becomes R = v0^2 sin(2θ)/g. This is the expression for the horizontal range with air resistance neglected and a landing height equal to the launch height. It also shows why the range is maximized at θ = 45°.

The key idea is that the horizontal range comes from multiplying the horizontal speed by the time the projectile is in the air. The horizontal speed is v0 cos θ, and for launch from ground level to land at the same level, the total flight time is T = 2 v0 sin θ / g. So the range is R = (v0 cos θ) × (2 v0 sin θ / g) = (2 v0^2 sin θ cos θ)/g. Using the identity sin 2θ = 2 sin θ cos θ, this becomes R = v0^2 sin(2θ)/g. This is the expression for the horizontal range with air resistance neglected and a landing height equal to the launch height. It also shows why the range is maximized at θ = 45°.