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The Physics of Throwing

Last modified 21Jan03 by Tony Dziepak

Here I describe the basic relationship between flight time, angle and height of release, velocities, and distance thrown. This applies with accuracy to the shot and hammer, where there are no significant aerodynamic forces. The areodynamic considerations are very significant in the discus and javelin throws. For those, please refer to Dyson's "Mechanics of Athletics."

A caveat: this formula may point to a particular optimal angle of release. However, we assume that the athlete can impart equal terminal velocity to the implement regardless of the release angle. If, however, the athlete can get a higher velocity at a lower angle, then we have a tradeoff between optimum angle and maximum release velocity.

I thought of this question when throwing both shot and hammer. In shot, it "feels" harder to put a shot vertically than horizontally because you are working against gravity. The difference probably differs across individual athletes. In hammer, especially short women, vertical angle may be limited by the need to prevent the hammer head from hitting the circle at the release orbit low point. Also, too steep an angle in the weight , esp. in initial winds, can interfere with smooth acceleration.


Variables I will use:

h: height of release (in meters) is the distance from the ground that the implement leaves the thrower. Roughly shoulder height for the hammer, and roughly body height for the shot.

d: distance thrown (in meters). I assume that the distance thrown is equal to the measuring distance (assume that the thrower releases the implement exactly above the origin of measurement.

V: velocity of implement at release (in meters per second), which can be separated into its horizontal and vertical components: Vx and Vy, respectively.

n: angle of release (in degrees).

And finally, t is a time measurement (in seconds). t=0 at moment of release.

Acceleration of the implement, while in flight, is always -9.8 meters/second squared; the act of gravity on the implement. Gravity is always acting vertically; there is no horizontal deceleration in the absence of aerodynamic forces. Acceleration is the same regardless of the weight of the implement.

The horizontal velocity is constant while the implement is in flight. It is Vcos(n).

If acceleration is always -9.8m/s^2, we can describe the vertical velocity of the implement at any time, t, as: Vy(t)=Vsin(n)-9.8t. In other words, the vertical velocity decreases by 9.8 m/s every second.

The vertical position (y) of the implement at any time can be described as the integral of the velocity equation. This is: y=h+V0yt-9.8/2(t^2).

The horizontal position is x=Vx(t).

Now I want to describe the relationship between angle of release, release velocity, and distance thrown. First, we can solve for the time, t', at which the implement lands. Upon landing, the vertical position is zero. Substituting, set 4.9t^2-Vsin(n)t-h=0, and solve for t. Using the quadratic formula, t'=(Vsin(n)+sqrt[(Vsin(n))^2 + 19.6h])/9.8.

Distance thrown, d=Vcos(n)t'; substitute:

d=Vcos(n)(Vsin(n)+sqrt[(Vsin(n))^2 + 19.6h])/9.8.

So this is the relationship of distance thrown (d), angle of release (n), height of release (h), and release velocity (V).

Some observations: the optimum angle of landing (assuming a flat ground, and no aerodynamic factors) is always 45 degrees. At a landing angle of 45 degrees, you get the maximum distance for a given release velocity.

However, because the implement is released at a height above the ground, the optimum angle at release is somewhat less than 45 degrees. For shot put, it is 40-42 degrees, and for hammer, it is 43-44 degrees.


Question: if a female throws a middle school shot (8 pounds, 3.629) 35' 6" (10.82 meters), how far could she throw a high school shot (4.00 kg)?

We cannot use this formula to determine this. We know that force equals mass times acceleration. However, with a heavier shot, assuming the thrower moves in the same body position, the time of applied force is longer because the length of the acceleration path (during the glide or spin) is constant.

From the formula, assuming the height of release is 1.7 meters and the angle of release is 41 degrees, the initial velocity is 13.47 meters/sec, and the momentum, P=mass time velocity, is 48.874.

If we generate the same momentum with a heavier shot, the initial velocity would be 12.22 meters/sec, resulting in a distance of 9.15 meters, or 30 feet 1/4 inch. However, because the shot is accelerated slower through the spin or glide, force can be exerted on the shot longer over the same, fixed acceleration path. The athlete does more work, and generates more release velocity.

The actual difference depends upon the athlete's technique and proficiency. My personal observations is that the athlete should be able to throw 32-33' with a 4K shot.


AERODYNAMIC EVENTS are more complicated (discus and javelin).

The best reference book for throwing physics is "Mechanics of Athletics" by Geoffrey Dyson.

Thanks to Kyle Berry for important comments.

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