Q.For a projectile launched and landing at the same height, with air resistance neglected, the angle of projection that gives the maximum horizontal range is:
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Start your 14-day free trial to unlock the full solution →For a projectile launched and landing at the same height with no air resistance, the angle that maximizes horizontal range is 45°.
Why this matters: the physics of projectile range
When you throw a ball, javelin, or shot put, the distance it travels depends on three things: how fast you release it, the angle at which you release it, and the height from which you release it. In the idealized case where the projectile lands at the same height it was launched—imagine throwing across a flat field—there is one angle that squeezes the maximum horizontal distance out of a given release speed.
The range of a projectile is given by
R = (v² sin(2θ))/g
where v is the release speed, θ is the angle of projection above the horizontal, and g is gravitational acceleration.
Notice that the range depends on sin(2θ). The sine function reaches its maximum value of 1 when its argument is 90°. So sin(2θ) is maximized when 2θ = 90°, which means θ = 45°.
At this angle, the projectile spends equal time climbing and descending, and the horizontal and vertical components of velocity are balanced in a way that maximizes the product of flight time and horizontal speed. Launch too flat and the projectile hits the ground quickly; launch too steep and it spends time going up and down rather than forward. Forty-five degrees is the sweet spot.
In real throwing events—shot put, javelin, discus—the optimum angle is less than 45°, typically 35–40°, because the athlete releases the implement from shoulder height, well above the landing surface. The asymmetry between release and landing heights shifts the optimum angle downward.
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