A projectile is launched at 45 degrees with a speed of 20 m/s. Neglecting air resistance, what is the horizontal range?

A projectile is launched at 45 degrees with a speed of 20 m/s. Neglecting air resistance, what is the horizontal range?

• A. 20 m
• B. 40.0 m
• C. 60.0 m
• D. 40.8 m

Answer: (D)

The range of a projectile on level ground (without air resistance) is determined by how far it travels horizontally during the time it’s in the air. The standard result is R = v^2 sin(2θ) / g. Here, the speed is 20 m/s and the launch angle is 45°. The sine term is sin(2θ) = sin(90°) = 1, so R = (20)^2 / g = 400 / 9.8 ≈ 40.8 m. You can also see this by splitting into horizontal and vertical parts: the horizontal speed is v cosθ = 20 cos45° ≈ 14.14 m/s, and the time of flight is T = (2 v sinθ)/g ≈ (2×20×0.7071)/9.8 ≈ 2.89 s. Then R = v_x × T ≈ 14.14 × 2.89 ≈ 40.8 m. The result is about 40.8 meters, which makes sense because at 45° you get the maximum range for a given speed (sin(2θ) is maximized at 2θ = 90°).

The range of a projectile on level ground (without air resistance) is determined by how far it travels horizontally during the time it’s in the air. The standard result is R = v^2 sin(2θ) / g.

Here, the speed is 20 m/s and the launch angle is 45°. The sine term is sin(2θ) = sin(90°) = 1, so R = (20)^2 / g = 400 / 9.8 ≈ 40.8 m.

You can also see this by splitting into horizontal and vertical parts: the horizontal speed is v cosθ = 20 cos45° ≈ 14.14 m/s, and the time of flight is T = (2 v sinθ)/g ≈ (2×20×0.7071)/9.8 ≈ 2.89 s. Then R = v_x × T ≈ 14.14 × 2.89 ≈ 40.8 m.

The result is about 40.8 meters, which makes sense because at 45° you get the maximum range for a given speed (sin(2θ) is maximized at 2θ = 90°).