A projectile is launched with speed 20 m/s at 30 degrees above the horizontal. What is its horizontal range?

A projectile is launched with speed 20 m/s at 30 degrees above the horizontal. What is its horizontal range?

• A. About 20.0 m
• B. About 35.4 m
• C. About 40.5 m
• D. About 50.0 m

Answer: (B)

For a projectile launched from and landing at the same height, the horizontal range depends on the initial speed and launch angle and is given by R = v^2 sin(2θ) / g. Here, v = 20 m/s and θ = 30°, so 2θ = 60° and sin(60°) = √3/2 ≈ 0.866. Compute R = 400 × 0.866 / 9.8 ≈ 346.4 / 9.8 ≈ 35.4 m. So the horizontal range is about 35.4 meters. This result follows from splitting the motion into horizontal and vertical parts: horizontal velocity is v cosθ, and the flight time for level ground is T = 2 v sinθ / g, giving R = v cosθ × T = v^2 sin(2θ) / g. If the launch and landing heights differed, the formula would be more complex, but for level ground this is the correct range.

For a projectile launched from and landing at the same height, the horizontal range depends on the initial speed and launch angle and is given by R = v^2 sin(2θ) / g. Here, v = 20 m/s and θ = 30°, so 2θ = 60° and sin(60°) = √3/2 ≈ 0.866. Compute R = 400 × 0.866 / 9.8 ≈ 346.4 / 9.8 ≈ 35.4 m. So the horizontal range is about 35.4 meters. This result follows from splitting the motion into horizontal and vertical parts: horizontal velocity is v cosθ, and the flight time for level ground is T = 2 v sinθ / g, giving R = v cosθ × T = v^2 sin(2θ) / g. If the launch and landing heights differed, the formula would be more complex, but for level ground this is the correct range.