Q.The optimum angle of release in the shot put is LESS than 45° mainly because:

For a projectile launched and landing at the same height, the range is

R = (v² sin(2θ))/g

and this is greatest when sin(2θ) = 1, i.e. at θ = 45^°. That result depends entirely on the assumption that the launch point and the landing point are level.

In the shot put they are not level. The shot leaves the hand at about shoulder height — roughly two metres above the ground on which it lands. The shot therefore begins its flight already "part-way up", so it does not need as much of its velocity spent going upward; more of that velocity can be spent going forward. The mathematics of a projectile released above the landing surface gives an optimum angle below 45° (in practice around 36°–40° for the shot put), and the greater the release height relative to the range, the further below 45° the optimum falls. …

Concept understanding: where the 45° rule comes from — and what it assumes

R is largest when sin(2θ) is largest. The maximum value of the sine function is 1, which occurs when 2θ = 90^°, i.e.

θ = 45^°

Why the shot put breaks the assumption

Picture the athlete at the moment of release. The shot leaves the fingers of a fully extended arm, above the shoulder — roughly 2 metres above the circle. It then lands on the ground, at height zero.

So the projectile is released above its landing surface. Two consequences follow:

1. Free flight time is a gift. Because the shot starts high, gravity has extra vertical distance to work through before the shot lands. The shot stays in the air longer than a level-to-level throw at the same angle would.
2. Vertical velocity is worth less; horizontal velocity is worth more. In a level throw, the only way to buy flight time is to send the shot upward. Here, some of the flight time is already paid for by the release height. So it becomes more profitable to spend the release speed on horizontal velocity than on vertical velocity — and that means lowering the angle.

The taller the release point (relative to the distance thrown), the further the optimum angle falls below 45^°. For the shot put — a comparatively short throw released from a comparatively high point — the optimum settles at roughly 36°–40°, which is why coaches teach an angle "a little under 45°".

Why the other options are wrong

(a) "The shot is very light and is affected strongly by air resistance."

This is factually backwards. The shot is a solid metal sphere — 7.26 kg for senior men, 4 kg for senior women — small in surface area and very dense. It has a very high mass-to-surface-area ratio, so aerodynamic forces are almost negligible compared with its weight. Precisely because of this, the shot is the closest thing in athletics to an ideal projectile. Air resistance is the reason the javelin's angle is low, not the shot's.

(c) "Gravity acts more strongly on heavy objects."

Gravity exerts a larger force on a larger mass, yes — but the resulting acceleration is the same for all masses (g ≈ 9.8 m/s²), because the larger force acts on a proportionally larger mass. A heavy shot and a light ball released together fall together (ignoring air). Mass does not appear anywhere in R = v^2sin(2θ)/g. So mass cannot be the reason the angle changes.

(d) "The athlete must throw the shot, not push it."

This is doubly wrong. First, the rule is the opposite — the shot must be put (pushed from the shoulder/neck), not thrown or slung. Second, and more importantly, a rule of the competition cannot be the biomechanical reason for an optimum angle. It affects technique, not the physics of the flight path. …