# Kicking a ball off a spherical rock (without hitting it on the way down )

1. ### teddy75

5
1. The problem statement, all variables and given/known data
A person standing at the top of a hemispherical rock of radius R kicks a ball (initially at rest) to give it an initial horizontal velocity v0 What is the minimum initial speed for the ball not to hit the rock on its way down?

2. Relevant equations
$$x=v_0 t$$
$$y=R-\frac{1}{2}g t^2$$

3. The attempt at a solution
$$v_0^2t^2+R^2-2R\left(\frac{1}{2}gt^2\right)+\frac{1}{4}g^2t^4=R^2+2Rd+d^2$$
$$v_0^2t^2+R^2-Rgt^2+\frac{1}{4}g^2t^4=R^2+2Rd+d^2$$
$$v_0^2t^2-Rgt^2+\frac{1}{4}g^2t^4=2Rd+d^2$$
Taking the derivative of d with respect to v0
$$2v_0t^2=2R\frac{dd}{dv_0}+2d\frac{dd}{dv_0}$$
$$0=\frac{dd}{dv_0}}{d}$$
$$0=\frac{v_0t^2-R}{d}$$
$$v_0t^2=R$$

Yeah... somehow I managed to screw that up so badly that not only did I get an equation with a t2 that I don't know what to do with, but I also got the units to not match up...
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. ### Delphi51

3,410
Use your first 2 equations directly. Set y = 0 and x = R which is the landing point you want. Solve one equation for time and sub into the other.

3. ### teddy75

5
Yes, but does that guarantee that the ball will not hit anywhere on the sphere on the way down?

4. ### DaveC426913

16,541
It does that by setting x = R. i.e. the x distance of the kick enables the ball to clear the radius of the hemisphere.

5. ### Delphi51

3,410
That is an interesting question, teddy. I'm not convinced it is obvious. I guess I would want to work out the distance from the center of the hill
squareRoot(x^2 + y^2) and make sure it is greater than R for all times between 0 and R/v. If you are doing this, note that you can use v = sqrt(.5*g*R) but you can't use t = R/v because that is only true at the endpoint. I'm getting a 4th degree expression in t . . . wish I had a number for R so I could graph it.

6. ### teddy75

5
With the person standing on top of the rounded part of the hemisphere and the flat part on the ground? Like something along the lines of this attempt at a diagram?
_
`\
----\

7. ### Delphi51

3,410
Oh, oh, it DOES hit the hemisphere! Back to the drawing board on this one!

8. ### teddy75

5
Yeah, I've been spending the past couple hours doing that... I keep getting time in there and then, since it's not given or required, I have to find a way to get rid of it. I tried taking the derivative after doing that also, and that just seems to mess things up even more. I also tried taking it as (d+R)2, where d is the distance between the hemisphere and the location of the ball...

9. ### Delphi51

3,410
Dave, I think it hits because
x^2 + y^2 = v^2*t^2 + (R - .5gt^2)^2 and with v^2 = .5gR
= R^2 - .5gRt^2 + .25gt^4
The question is, is -.5gRt^2 + .25gt^4 > 0 for the time range?
This expression is .25gt^2[t^2 - 2R]
and it is clearly negative for small times.
Perhaps Teddy's approach using these expressions will yield a "safe" velocity. I'm going to study his work.

10. ### RoyalCat

671
I think the easiest way to approach this would be using analytical geometry. Just demand that the graph of the parabola $$y(x)$$ (Develop this as you would for projectile motion) is always greater than the equation representing the spherical rock.

In other words, demand that the function describing the difference between the two, is always positive. The solution is really quite elegant.

Last edited: Oct 7, 2009
11. ### Delphi51

3,410
That sounds good, RoyalCat. But my geometry is weak!

Teddy, I don't think you need to differentiate. Or add the d to R. Just write that the ball's distance from the center must be > R.
Avert your eyes if you don't want to be given the solution!

Just x^2 + y^2 > R^2 gives
v^2*t^2 - Rgt^2 + .25g^2*t^4 > 0
v^2 - Rg + .25g^2*t^2 > 0
V^2 > Rg - .25*g^2*t^2
This cries out for a solution of v = sqrt(R*g).
That works for all times > 0, doesn't it? And of course it has to touch at t = 0. I don't think it could be any less because you could always choose a small enough t so the t^2 term would be practically zero.

12. ### RoyalCat

671
Doing that is just mathematically unsound, though. You must then demand that your expression holds true for all $$t$$ during the flight-time. You have to look at the end points of $$t=0$$ AS WELL AS $$t=\frac{R}{v_0}$$ which to me, seems far too complicated.

Instead of giving up at the last step, think of what the parabola represents. It is the difference between the spherical surface and the projectile, as a function of time. How would you ensure that it is always greater than 0?

13. ### teddy75

5
That was actually my original approach, but how did you just drop the t2 in the final solution?

14. ### Delphi51

3,410
V^2 > Rg - .25*g^2*t^2
I definitely considered all times t > 0.
Certainly V^2 = Rg is large enough, because the t^2 term is always positive for t > 0, always taking something away from the Rg on the other side and ensuring the inequality is true.
It is not unnecessarily large because in the limit as t approaches zero the t^2 term is made as small as you can imagine. Thus any decrease to the V^2 = Rg solution (which is independent of time) would have to be infinitesimally small in order to satisfy the inequality as t --> 0.

15. ### RoyalCat

671
Oh, I see now. But you're completely over-complicating everything.
It is completely obvious (Once you stare at the inequality for a couple of seconds) that $$v_0^2>Rg$$ is the solution.
You threw in some limits and differential calculus where you need none, and where it only obfuscates the conclusion.

The term in $$t^2$$ is always negative. If we choose $$v_0^2$$ so as $$v_0^2>Rg$$ then our LHS is greater than $$Rg$$ for all values of $$t$$ (It is a constant.)
On the other hand, as time progresses, the RHS becomes more and more negative, increasingly smaller than $$Rg$$
From these two facts it immediately follows that $$v_0>Rg$$ satisfies the inequality for all $$t$$

That is elegant reasoning indeed. :) Too bad you got tangled up with limits and whatnot.

The route I chose with this problem was the grade-school one. I just looked at the parabola and said that I want it to always be positive, so I took the discriminant and looked at when it is negative, and got the $$v_0^2>Rg$$ solution.

Last edited: Oct 7, 2009
16. ### Delphi51

3,410
Ah, you mean look at the graph of -.25*g^2*t^2 and see that it peaks at t = 0 so is negative for all other values! Yes, I see that now - no need to use the limit argument.

Or perhaps you mean look at the distance squared
v^2*t^2 - Rgt^2 + .25g^2t^4 + R^2 to see what values of v ensure it is greater than R^2. With T = t^2 it becomes
y = .25g^2*T^2 + (v^2 - Rg)T + R^2, a quadratic
with its vertex at T = -b/(2a) = (Rg-v^2)/(.5g^2)
so if v^2 = Rg the vertex is at T = 0 and at all other values of T the distance expression will be greater than R^2. If the vertex is to the right of T = 0 it won't do so v^2 must be greater than or equal to Rg.

Well, this analysis is most interesting and adds another level of delight to the whole problem. Thank you!

Hats off to the prof who came up with the question. I fear that in my enthusiasm I may have spoiled the challenge for the student(s) who were watching it last night.