Originally posted by forkedknight
a 45 degree angle would minimize initial speed, and I would guess in turn that would minimize the average speed.
The average speed = (arc length) / (time)
solving for either involves integrating and/or deriving, and I don't really feel like setting that one up.
Starting with the force balance on the ball in the time domain:
x" = 0
y" = -m.g
We can derive equations for the velocity:
x' = v.cos(theta)
y' = v.sin(theta) - m.g
x = v.t.cos(theta)
y = v.t.sin(theta) - (1/2).m.g.t^2
Now, the angle of the toss and the velocity have to match up, because not every velocity will cause the ball to hit the target given a specific toss angle. Using the fact that y(0) = y(L) = 0, we find that:
v = SQRT(m.g.L/sin(2.theta))
Using this result to convert from the time domain to the distance domain, we get a surprisingly simple result:
y = tan(theta).(x - x^2/L)
The formula for arc length is given by:
A = int(0,L) (SQRT(1 + (dy/dx)^2)) dx
Where "int(0,L)" is the integral from x=0 to x=L. Going back to the time domain for a moment, and using the fact that x(t-final) = L, we can find the total time for the ball to reach the target:
t-final = L/(v.cos(theta)) = L/(cos(theta).SQRT(m.g.L/sin(2.theta)))
To find the average speed, we divide the arc length by the total time:
avg. speed = A / t-final = (int(0,L) (SQRT(1 + (dy/dx)^2)) dx) / L/(v.cos(theta))
I thought there might be a shortcut that allows us to get around having to solve the integral for the arc length by differentiating (A / t-final) right off the bat, but it seems to come down to a differentiation using the product rule and the ugly integral persists.
So I took my own shortcut! I plugged the numbers into Excel, and graphed the parabola using 101 points. Then I used Excel Solver to find the minimum average speed. The actual speed was different depending on the values used, but the angle always came out to 62.8 degrees. If anyone is brave enough to solve this analytically, it would be interesting to see the solution. 🙂