Finding Min Velocity for Projectile from A to B

[ed_uk]

Homework Statement

What I would like to do is find the minimum velocity needed to launch a projectile from point A to B when B is not at the same height as A. I keep finding myself stuck with two unknowns (initial speed and release angle) and I am not sure how to resolve them. If the problem is simplified and the angle is fixed or the initial speed is fixed then the solution resolves to one variable and is relatively easy to solve. However if release angle or initial speed is not fixed then there is an infinite number of angles and speeds which will take a projectile from A to B. As the release angle approaches 90 degrees the initial speed tends towards infinity as it does if the angle "points" directly at the target. So I assume there is a middle ground where the minimum velocity can be calculated. Any help is much appreciated, many thanks.

Homework Equations

I'm not sure how to paste equations onto this page but most of the information I have found is on http://en.wikipedia.org/wiki/Range_of_a_projectile and http://en.wikipedia.org/wiki/Projectile_motion.

The Attempt at a Solution

If point A and B are flat then the "perfect" angle is 45 degrees and the speed needed to get from A to B is quite easy to work out. However if B is above or below A then I'm not sure how to calculate the "perfect" angle.

 

[mathguy2009]

Hi ed_uk!

I'm assuming you've had exposure to
a) trigonometry
b) calculus

Intuitively speaking, it seems that the velocity is inextricably linked to the angle. If you think about throwing a ball at a target at a fixed angle, there is only one speed that will get the ball there. However, if you are throwing a ball at a target with a fixed velocity, then it doesn't necessarily mean there is only one angle that will get the ball there. Thus there seems to be a sort of conditional logic governing the situation here:

If angle --> then velocity​

So the velocity is linked to the angle, but not necessarily the other way around. Based on this, I think if you find the angle, you'll find the velocity. This almost assuredly has some relation to trigonometric functions, such as y = sin x, in which you can get y if you know x, but if you know y you obtain a set of x's (strikingly similar to the logic mentioned above!). Because of this, I suspect that you need to produce a formula such that the velocity is the dependent variable and the angle is the independent variable which should be found within a trigonometric function of some sort.

In addition, this is an extrema problem (remember those from calculus?), so after finding the formula, simply take a derivative, set equal to zero, and solve.

I'm not sure myself concerning what the formula is yet, but I'm working on it. Hope this helps!

 

[Libohove90]

Well I derived this equation:

v_{}0 = \Deltax \sqrt{}g / -2 (cos \phi)^2 (\Deltay - x tan\phi)

With this equation you can pick any \Deltay as long as you know \Deltax and \phi.

 

[jaumzaum]

Well, we know that

<br /> Voy = V.sen\alpha

Vx = V.cos\alpha

Vy = Voy - gt = V.sen\alpha - gt

\Delta y = Voy.t - gt²/2 = V.sen\alpha - gt²/2

\Delta x = Vx.t = V.cos\alpha .t -&gt; t = \Delta x/V.cos\alpha\<br />


Replacing in the y equation:
<br /> \Delta y = \Delta x.tan\alpha - g\Delta x²/2V²cos ²\alpha<br /> <br />

A little bit algebra
<br /> <br /> V=\sqrt{ \frac{g\Delta x²}{2cos²\alpha (\Delta x tan \alpha - \Delta y)}} = \sqrt{g/2} \frac{\Delta x} { \sqrt{\Delta x sen _{2\alpha} - \Delta y ( cos _{2\alpha} +1)}}<br />


For V min we have to have \Delta x sen_{2\alpha} - \Delta y ( cos _{2\alpha} +1) max

y= \Delta x sen_{2\alpha} - \Delta y ( cos _{2\alpha} +1)

y&#039;= \Delta x cos_{2\alpha} + \Delta y sen _{2\alpha}

\Delta x \sqrt{1-sen² _{2\alpha}} + \Delta y sen _{2\alpha} = 0

sen _{2\alpha} = \Delta x /\sqrt{\Delta x² + \Delta y ²}

cos _{2\alpha} = \Delta y /\sqrt{\Delta x² + \Delta y ²}




We see that 2\alpha is the angle formed by the initial and final position of the ball and the vertical rect.
So 2\alpha=\beta




If we replace we get \sqrt{g/2} \frac{\Delta x} { \sqrt{\Delta x sen \beta - \Delta y ( cos \beta +1)}} = \sqrt{g/2} \frac{\Delta x} { \sqrt{ \Delta x cos \theta- \Delta y ( sen \theta +1)}}<br />

 

[ed_uk]

wow thanks for the fast replies.

Firstly mathguy

However, if you are throwing a ball at a target with a fixed velocity, then it doesn't necessarily mean there is only one angle that will get the ball there

That's right, there are two possible angles. The larger the velocity the further apart those angles are and the lower the velocity the closer those two angles are. So there is a point where "both" angles are the same, which would be the "perfect" angle.

In addition, this is an extrema problem (remember those from calculus?), so after finding the formula, simply take a derivative, set equal to zero, and solve.

I dimly remember derivatives. I thought they found the rate of change of a curve? So I could use them to find the maximum height of a curve, knowing the rate of change is 0, but I'm not sure what curve to plot where zero would mean I had found my "perfect" angle/velocity. Thanks for the ideas though.

Libohove

With this equation you can pick any Δy as long as you know Δx and ϕ.

The problem is that I don't know what ϕ should be.

Finally jaumzaum

I'm sorry I can't quite follow all the algebra. We still seem to be left with equations with two unknowns? I'm not sure how I could find the minimum velocity (velocity being initial speed and angle of release) using those equations?

Many thanks to you all for replying.

 

[ed_uk]

I have been thinking a little more.

I said before that there are two possible angles depeding on the initial speed and that these angles converge. If I plot a graph against initial speed with the difference of the two angles I get a graph which goes from 90 to 0 with the turning points at 90 and 0 and could therefore find the derivative and solve this to find initial speed. I think? I'll see if I can work this out.

thanks

 

[jaumzaum]

I'm sorry I was not clear

Finally jaumzaum



I'm sorry I can't quite follow all the algebra. We still seem to be left with equations with two unknowns? I'm not sure how I could find the minimum velocity (velocity being initial speed and angle of release) using those equations?

\theta ANGLE is is the angle formed by the initial and the final position of the target (that we know!) and the horizontal rect
\beta angle is the angle formed by the initial and the fnl position of the target and the vertical rect
\alpha angle is the angle that we had to discover, a unknown, but we found out that it is half of Beta, that is complementary to Alpha!




We have the initial and final position, so we have sin/cos and tan Alpha, Beta, and sin Theta we can calculate from the formule \sqrt{(1-cos\beta)/2}



[]'s
John

 

[ed_uk]

John,

Thank you very much. Took me a few read-throughs but I get it now :) You've been a great help that is exactly what I was looking for and it all seems to work out.

p.s. you were plenty clear enough just took me a while to understand the notation, it's been a while since I did calculus or algebra.

Many thanks