# Softest toss

GregM
Posers and Puzzles 01 May '08 18:56
1. 01 May '08 18:56
Suppose you toss an object from one point to another point with the same altitude; under the influence of gravity, it follows a parabolic trajectory. What trajectory minimizes the average speed of the object? The obvious guess is probably the right answer, but I haven't proven it.

What if the starting and ending points don't have the same altitude?
2. wolfgang59
Mr. Wolf
01 May '08 21:17
Originally posted by GregM
Suppose you toss an object from one point to another point with the same altitude; under the influence of gravity, it follows a parabolic trajectory. What trajectory minimizes the average speed of the object? The obvious guess is probably the right answer, but I haven't proven it.

What if the starting and ending points don't have the same altitude?
Lets be clear and assume you mean VELOCITY.

Do you really mean the AVERAGE velocity in flight or the velociy at impact?
3. 01 May '08 21:392 edits
Originally posted by wolfgang59
Lets be clear and assume you mean VELOCITY.

Do you really mean the AVERAGE velocity in flight or the velociy at impact?
What's wrong with the question as posed? The average speed (i.e. the average magnitude of velocity).

After all, in the symmetric case, if you want the average velocity you can ignore the vertical motion as these will cancel out. And the horizontal velocity is constant (ignoring air resistance etc). Whereas minimizing the average speed might be a more interesting question. I'll work it out later.
4. forkedknight
Defend the Universe
02 May '08 00:06
Originally posted by mtthw
What's wrong with the question as posed? The average speed (i.e. the average magnitude of velocity).

After all, in the symmetric case, if you want the average velocity you can ignore the vertical motion as these will cancel out. And the horizontal velocity is constant (ignoring air resistance etc). Whereas minimizing the average speed might be a more interesting question. I'll work it out later.
agree with mattw

minimizing average velocity would involve throwing the object as high as possible, so it will take the longest amount of time to reach its destination.
5. 02 May '08 00:231 edit
Indeed, I mean average speed; average velocity is simply the horizontal component of initial velocity. This is a more interesting question as neither of the extremes can be correct (throw the ball with huge speed either nearly horizontally or nearly vertically).
6. forkedknight
Defend the Universe
02 May '08 00:321 edit
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.
7. wolfgang59
Mr. Wolf
02 May '08 07:52
Originally posted by mtthw
What's wrong with the question as posed? The average speed (i.e. the average magnitude of velocity).

After all, in the symmetric case, if you want the average velocity you can ignore the vertical motion as these will cancel out. And the horizontal velocity is constant (ignoring air resistance etc). Whereas minimizing the average speed might be a more interesting question. I'll work it out later.
ðŸ˜³

8. PBE6
Bananarama
02 May '08 17:54
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

And position:

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. ðŸ™‚
9. 03 May '08 07:54
Originally posted by PBE6
the angle always came out to 62.8 degrees.
Really? Not 45? Cool, I was hoping for a non-trivial answer.
10. 03 May '08 11:48
11. 03 May '08 12:38
Originally posted by wolfgang59
Lets be clear and assume you mean VELOCITY.

Do you really mean the AVERAGE velocity in flight or the velociy at impact?
Average velocity for a throw with same altitude at beginning and end would simply be the horizontal component of the throw.

So I assume that you don't know much about this sort of question and were just grammar mongering, which you had previously indicated you disliked. By the way, in your post, velocity is spelled with a 't' EVERY time, not just the first instance.
12. 03 May '08 22:26
@pbe6

great job, you have done!