# Minimum Velocity needed to clear a hemisphere

1. Feb 8, 2010

### simpleton

1. The problem statement, all variables and given/known data

There is a hemisphere of radius R metres on the ground. You are standing at the very top of the hemisphere and you kick a ball out such that it travels outwards horizontally with speed v metres per second. During the trajectory of the ball, it does not come into contact with the hemisphere at all.
a) What is the minimum v that can fulfil the above conditions?
b) How far away from the centre of the hemisphere will the ball land?

2. Relevant equations
Projectile Motion Equations

3. The attempt at a solution

Looking at the y-axis : vt = L, where L is the horizontal distance travelled after time t
Looking at the y-axis : H = 0.5*g*t^2, where H is the vertical distance from the top of the hemisphere after time t

Therefore, at time t, the distance of the ball above the ground is R - 0.5*g*t^2.

At the point that is x metres away from the centre, the height of the slice of hemisphere at that point is sqrt(R^2 - x^2).

I can substitute x with vt and get sqrt(R^2 - v^2*t^2)

So now, in order to fulfil the conditions:

R - 0.5*g*t^2 > sqrt(R^2 - v^2*t^2) for t = 0 to the time the ball reaches the ground

Squaring both sides, I get:

R^2 - R*g^2*t^2 + 0.25*g^2*t^4 > R^2 - v^2*t^2 (I think I dont have to change the bigger-than sign to smaller-than because I know that for this range of values, both sides>0)

So if I make v the subject:

v^2*t^2 > R^2 + R*g^2*t^2 - 0.25*g^2*t^4
v^2 > R^2 + R*g^2*t^2 - 0.25*g^2*t^4/t^2
v > sqrt(R^2 + R*g^2*t^2 - 0.25*g^2*t^4)/t

But I still have the t term on the Right Hand Side. How do I get rid of it, or is there another way to do this question?

2. Feb 11, 2010

### simpleton

3. Feb 11, 2010

### ehild

there is a mistake: g is not squared:

R^2 - R*g*t^2 + 0.25*g^2*t^4 > R^2 - v^2*t^2

You can subtract R^2 from both sides:

- R*g*t^2 + 0.25*g^2*t^4 > - v^2*t^2

The inequality has to hold for all t, even at t=0. So v^2>R*g. If v is less, the ball rolls downhill.

ehild

4. Feb 11, 2010

### simpleton

Thank you very much! :)