In projectile motion on level ground, the maximum horizontal range occurs at which launch angle (neglect air resistance)?

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Multiple Choice

In projectile motion on level ground, the maximum horizontal range occurs at which launch angle (neglect air resistance)?

Explanation:
In projectile motion on level ground without air resistance, the horizontal range depends on the angle through the expression R = v0^2 sin(2θ)/g. The only angle-dependent part is sin(2θ), which is maximized when 2θ = 90°, so θ = 45°. That angle balances vertical rise and horizontal speed to give the longest flight time and the farthest landing distance. Angles like 0 degrees produce no vertical lift and therefore zero range, while 30 and 60 degrees give sin(60°) ≈ 0.866 (or sin(120°) ≈ 0.866), which is less than the maximum. Hence 45 degrees yields the maximum horizontal range.

In projectile motion on level ground without air resistance, the horizontal range depends on the angle through the expression R = v0^2 sin(2θ)/g. The only angle-dependent part is sin(2θ), which is maximized when 2θ = 90°, so θ = 45°. That angle balances vertical rise and horizontal speed to give the longest flight time and the farthest landing distance. Angles like 0 degrees produce no vertical lift and therefore zero range, while 30 and 60 degrees give sin(60°) ≈ 0.866 (or sin(120°) ≈ 0.866), which is less than the maximum. Hence 45 degrees yields the maximum horizontal range.

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