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 25^{2}*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.