Best Javelin Throw

Question:

I've been having trouble trying to set up this problem. An athlete consistently throws a javelin at a speed of 25 m/s. What is her best distance? Neglect friction.

Answer:

Let's define the distance of the throw to be the distance covered at the time the javelin returns to the altitude from which it was released. This definition avoids having to deal with the height of the release point. The best distance will result from the combination of maximum time in the air and best horizontal velocity. If the javelin went straight up, the time in the air would be maximum but the horizontal velocity would be zero. If the javelin went straight out, horizontal, the time in the air would be zero (under our definition) but the horizontal velocity would be 25 m/s, the maximum. The problem becomes one of optimizing the vertical velocity component to get the best combination of time of flight and speed over the ground.

Let's call the horizontal velocity Vx and the vertical velocity Vy. The angle of elevation of the flight path we will call E. With these definitions Vx=25 m/s * cos(E) and Vy=25 m/s * sin(E).

Neglecting air resistance the path of the javelin will be symmetrical about the highest point of its flight. The time (t) to reach that highest point will be the vertical velocity divided by the constant acceleration of gravity (9.8 m/s/s). t=25 m/s*sin(E)/9.8 m/s/s.

Due to the symmetry of the path the total flight time will be 2*t. The distance covered will be the horizontal velocity times the total flight time or 2*t*25*cos(E), or, 2*25/9.8*sin(E)*25*cos(E).

There is a trig identity that 2*sin(E)*cos(E)==sin(2*E) so the distance covered is 25²*sin(2*E)/9.8. This distance will be maximum when sin(2*E) takes on its maximum value of 1.0. That happens when 2*E=90 degrees or E=45 degrees. The best distance then will be the initial velocity squared divided by the acceleration of gravity. In this case about 63.8 meters.