How Fast Must a Ball Be Kicked to Avoid Hitting a Hemispherical Rock?

[Ailo]

Homework Statement

A person standing at the top of a hemispherical rock of radius R kicks a ball (initially at rest on the top of the rock) to give it horizontal velocity v_i
(a) What must be its minimum initial speed if the ball is never to hit the rock after it is kicked?
(b) With this initial speed, how far from the base of the rock does the ball hit the ground?

Homework Equations

\frac{1}{2}gt^2=R=y, \ v_it=x, \ (v^2/R=a_c ?)

The coordinate system I use is one with a horizontal x-axis in the direction of the kick and a vertical y-axis downwards.

The Attempt at a Solution

Basically, I just tried gathering as much information as possible. I got t=\sqrt{2R/G}. I also managed to link x and the initial speed in an equation: gx^2=2v_i^2R. But I'm stumped as to how I can assure that the ball doesn't hit the rock. Maybe I have to use what I know about circular motion as well, but that's just a wild guess... Any hints? =)

 

[Hootenanny]

Welcome to Physics Forums.

What is the vertical and horizontal distance from the point where the ball is kicked to the edge of the hemisphere in contact with the ground?

 

[Ailo]

Thank you. =)

R, and R.

 

[Hootenanny]

Thank you. =)

R, and R.

Good. So when y=R, what must be the value of x so that the ball doesn't touch the hemisphere?

 

[Ailo]

x must be greater than R.

 

[Hootenanny]

x must be greater than R.

Correct. So can you solve the problem?

 

[Ailo]

I only get v_it>R \Rightarrow v_i>R/t. Even if i subsitute in what I know for t, R/t can't be the minimum velocity. Or did you mean something else?

 

[Mentallic]

x must be greater than R.

My initial thought was this too. Projectile motion follows a parabolic path. Couldn't this path - if the ball were to kicked precisely enough to hit the edge of the rock on the ground - intersect with the rock at some point along the journey?
I'm merely saying this because by visual inspection, a circle 'drops off' quicker near the end compared to a parabola.

http://img294.imageshack.us/img294/6049/ballrockmm0.png
http://g.imageshack.us/img294/ballrockmm0.png/1/

Notice the intersection between the path of the ball and the rock face as the ball tries to land at the corner of the rock. This could suggest that the minimum horizontal velocity v_i needs to be increased/re-thought.

 

[Mentallic]

(b) With this initial speed, how far from the base of the rock does the ball hit the ground?

...And the second question supports my suspicions.

 

[Ailo]

Yes.

gx^2=2v_i^2y

 

[Hootenanny]

You make a good point Mentallic. We need to reformulate the problem.

Let us define our axes such that the the origin coincides with the centre point of the circle. The the ball's initial position is (x,y) = (0,R) so the equations of motion become:

y = R - \frac{1}{2}gt^2

x = v_i t

As Mentallic points out, we now also have an additional constraint required to ensure that the ball does not intersect the hemisphere:

x^2+y^2 > R^2 \;\;\;\text{ for }\;\;\; t>0

Do you follow?

 

[Ailo]

Yes, now I follow. =)

 

[Hootenanny]

Yes, now I follow. =)

Good. So the next step would be to find an expression for v_i in terms of t. Can you do that?

Spoiler

HINT: Substitute the first two equations into the inequality to eliminate x and y

 

[Ailo]

v_i^2t^2+(R-\frac{1}{2}gt^2)^2>R^2

After a lot of algebra, this reduces to

v_i^2>Rg-\frac{g^2t^2}{4}

So how do I find the t I'm after?

 

[Hootenanny]

v_i^2t^2+(R-\frac{1}{2}gt^2)^2>R^2

After a lot of algebra, this reduces to

v_i^2>Rg-\frac{g^2t^2}{4}

So how do I find the t I'm after?

Looking good so far. What does t represent in this case? In other words, what specific time are we looking at?

 

[Ailo]

When t=0, v_i>\sqrt{Rg} !

I think I actually understood that. A million thanks Hootenanny (and Mentallic also)! =)

 

[Ailo]

Now b is straightforward. When the ball hits the ground, y=0.

R-\frac{1}{2}gt^2=0 \ \Rightarrow \ t=\sqrt{\frac{2R}{g}}

The ball has then traveled the horizontal distance

v_it=\sqrt{Rg} \cdot \sqrt{\frac{2R}{g}}=\sqrt{2}R.

So the answer is (\sqrt{2}-1)R.

 

[Hootenanny]

When t=0, v_i>\sqrt{Rg} !

I think I actually understood that. A million thanks Hootenanny (and Mentallic also)! =)

Now b is straightforward. When the ball hits the ground, y=0.

R-\frac{1}{2}gt^2=0 \ \Rightarrow \ t=\sqrt{\frac{2R}{g}}

The ball has then traveled the horizontal distance

v_it=\sqrt{Rg} \cdot \sqrt{\frac{2R}{g}}=\sqrt{2}R.

So the answer is (\sqrt{2}-1)R.

Let's have a reality check here:

x = \left(\sqrt{2}-1\right) R \approx \left(1.414 - 1 \right) R = \left(0.414\right)R

Now does that satisfy the the constraint

x^2 + y^2 > R^2

 

[Ailo]

What I'm saying here is that the ball hits the ground a horizontal distance 1,414*R units from where it started. This satisfies the inequality. But b asked how far from the base of the rock the ball lands. Now, I don't have english as my mother tongue, but I assume that the distance from the base means the distance from the rock's lowest part.