# Hard projectile problem

1. Jan 17, 2009

### Ailo

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 on the top of the rock) to give it horizontal velocity vi
(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?

2. Relevant 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.

3. 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? =)

2. Jan 17, 2009

### Hootenanny

Staff Emeritus
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?

3. Jan 17, 2009

### Ailo

Thank you. =)

R, and R.

4. Jan 17, 2009

### Hootenanny

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

5. Jan 17, 2009

### Ailo

x must be greater than R.

6. Jan 17, 2009

### Hootenanny

Staff Emeritus
Correct. So can you solve the problem?

7. Jan 17, 2009

### 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?

8. Jan 17, 2009

### Mentallic

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.

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.

9. Jan 17, 2009

### Mentallic

...And the second question supports my suspicions.

10. Jan 17, 2009

### Ailo

Yes.

$$gx^2=2v_i^2y$$

11. Jan 17, 2009

### Hootenanny

Staff Emeritus
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?

12. Jan 17, 2009

### Ailo

Yes, now I follow. =)

13. Jan 17, 2009

### Hootenanny

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

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

Last edited: Jan 17, 2009
14. Jan 17, 2009

### 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?

15. Jan 17, 2009

### Hootenanny

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

16. Jan 17, 2009

### Ailo

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

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

17. Jan 17, 2009

### 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 travelled the horizontal distance

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

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

18. Jan 17, 2009

### Hootenanny

Staff Emeritus
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

$$x^2 + y^2 > R^2$$

19. Jan 17, 2009

### 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.