## Why is the Javelin not thrown at 45°?

The best way to explore this is to use paramteric equations.  First make sure you are set to DEGREES (not radians), and in 2D, enter x = (vcosb)t, y = h + (vsinb)t − ½gt² with startup options on “Manual”, with ‘t’ from 0 to 8 (sec), step 0.01.  Set the constants to: ‘v’ = 10 (m/s), ‘b’ = 45 (°). h = 1.8 (m), g = 9.81 (m/s²).
After plotting, select the curve and right-click “enter point on curve” at t = 0, with t-snap = 0.1, and plot teh Velocity Vector.

For 3D, enter x = (vcosb)(cosa)t, y =(vcosb)(sina)t, z=h+(vsinb)t − ½gt²   where ‘b’ is the angle to the horizontal (as in 2D), and ‘a’ is the angle round from the ‘x’ axis. Proceed as with 2D to place a point at t = 0 and a velocity vector.  Draw the plane z = 0 to represent the ground, and a number of circles on the grass using the form x = 10cost, y = 10sint, z = 0.

To investigate, either in 2D or 3D, note that the maximum orizontal range ooccuras at 45°, whereas the maximum range on the ground occurs around 39°. This analysis assumes that the javelin can be thrown at the same speed at all angles, which is clearly not the case. You can show that the range is much more affected by the initial velocity of throw than by the angle. Research elswhere suggests that the physiology of the human arm allows the maximum speed of projection at around 31°, which is the angle that most javelin throwers use.