Minimum initial velocity for throwing a ball off a hemispherical rock

[diredragon]

Homework Statement

At which minimum velocity should you throw the ball horizontally if you are standing on a hemispherical rock of radius R so that it at no point touches the rock and lands at the minimum distance from the rock horizontally. Find the expression that solves for initial velocity and woth that velocity calculate the distance traveled from the rock by the ball.

Homework Equations

X=Xi + Vcosu(t)
Y=Yi + Vsinu(t) - g/2 (t^2)
Vx=Vcosu
Vy=Vsinu - gt

The Attempt at a Solution

It is definatelly clear that yf= 0 and yi=R
From there t=x/V and 0=R-(1/2)gt^2 i gey R=(1/2)g(x^2/V^2), where to fo from here? x can't be R its path is oarabolic and would hit the rock if set on R and v is also unknown. It seems that there must be another correlation between y, x and R that i could use, but i don't see it.

 

[RUber]

Can you define a path tracing the rock, and then write y_{ball} - y_{rock} > 0 for t< t_final?

 

[RUber]

I get something that looks like this:
Rock is defined by $x^2 +y^2 =R^2$ which gives $y_{rock} = \sqrt{R^2 - x^2 }$
The path in x of the ball is defined by velocity and time, as in $x = vt$ initial trajectory is horizontal.
The path in y of the ball is defined by R and time. $y = R - 4.9t^2$.
You want to find a velocity such that the only solution to y - y_{rock} = 0 is t = 0 (or possibly the t_{final}).

 

[diredragon]

I used the y(ball)^2 + x^2 > R^2 and got to the equation i posted below. Should i now solve for v?

 

[RUber]

That should work. I don't see your equation, but I imagine it looks like:
$(R-4.9t^2)^2 + (vt)^2 > R^2$
Which should simplify down to something like
$t^2 ( t^2 + f(v)) > 0$ for all t, which means $f(v)$>0.
So, solve f(v) = 0 for the minimizer.

 

[diredragon]

Yes that is the equation but i don't know how you got f(v) there and what minimizer is.

 

[RUber]

Yes that is the equation but i don't know how you got f(v) there and what minimizer is.

f(v) is just a placeholder for some function that depends on v, in this case is should be f(v) = v^2 - 9.8R.
You are looking for the smallest velocity such that the inequality is true, so you are looking for the minimizer of the inequality.

 

[diredragon]

So its (gR)^(1/2) and the distance from the rock at the point of fall is 0 = R - (1/2)g(x^2/gR) so Δx= R( √2 + 1)

 

[RUber]

Where did the +1 in your distance for x come from?

 

[diredragon]

Where did the +1 in your distance for x come from?

It should be -1

 

[RUber]

I'm still not seeing how you are coming up with that. What are you solving and how are you solving it?

So its (gR)^(1/2) and the distance from the rock at the point of fall is 0 = R - (1/2)g(x^2/gR) so Δx= R( √2 + 1)

 

[diredragon]

I'm still not seeing how you are coming up with that. What are you solving and how are you solving it?

I want to know how far off the rock the ball is horizontally. I replace V^2 by gR in equation 0=R - 1/2g(x^2/gR) solve for x and calculate the Δx quantity to be Δx=x-R
From above x=√2R so √2R - R = R(√2 - 1)

 

[insightful]

Are we not looking for a parabola whose second derivative is the same at (0,R) as a circle centered at (0,0) with radius R?

This also gives the distance the ball hits from the rock as R(sqrt(2)-1).

 

[diredragon]

Are we not looking for a parabola whose second derivative is the same at (0,R) as a circle centered at (0,0) with radius R?

This also gives the distance the ball hits from the rock as R(sqrt(2)-1).

Could you explain a little further how exactly does second derivative of a parabola being the same as that of circle at the point (0,R) correlates to the problem. How does one find the function if he knows f''parabola = f''circle located at point (0,0) with radius R at point (0,R). I have never met anything similar so I am unfamiliar.

 

[RUber]

Oh, I see. The question was asking for the horizontal distance from the rock. Sorry for not understanding before.
I think you have this problem fully solved.
There are many other ways to attack this problem, one of which is what Insightful mentioned. However, you have been able to solve it with the tools you have.
Play around with the problem is a few ways, and you might find more relationships that will have to be true.

 

[insightful]

Could you explain a little further how exactly does second derivative of a parabola being the same as that of circle at the point (0,R) correlates to the problem. How does one find the function if he knows f''parabola = f''circle located at point (0,0) with radius R at point (0,R). I have never met anything similar so I am unfamiliar.

It's simply that the curvature of the parabola at the top of the dome must equal the curvature of the dome. The slope of each (f') is obviously zero at that point, so the rates at which the slopes are changing (f") being equal seemed to me a reasonable approach. I can't be more specific because it just came to me while sketching circles under parabolas.

Y₁=sqrt(R²-x²)
Y₂=-Ax²+R
Y₂'=-2Ax
Y₂"=-2A

so then get Y₁" (a little messy, so not shown here) and set equal to -2A at x=0 to get A=1/(2R) and
Y₂=-x²/(2R)+R as the equation of the parabola.