# Finding ymax and ymin of a ball thrown straight into the air

1. Oct 21, 2009

### babygiggles85

1. The problem statement, all variables and given/known data
A 300g ball is thrown straight up into the air. It is exactly 2m above the ground when it is released. It reaches a height of 12m above the height from which is was released, and then falls straight back down.

a) assume that y=0 is located at the point of release of the ball. Find the maximum and minimum values of y and PE(g). What is the total energy of the system? Find the maximum and minimum values of KE.

b) repeat part (a) with y=0 located on the ground

2. Relevant equations

I'm having trouble finding what equations to use. I tried using mgh + 1/2m(vf^2-vi^2) but the problem doesn't give velocity

3. The attempt at a solution

2. Oct 21, 2009

### Raziel2701

Here's the thing about velocity, when the ball reaches its maximum altitude, its velocity is zero. You might need to employ kinematic equations for this one. Also mind the direction of gravity. When the ball is going up, gravity is pulling down.

Regarding the energy of the system, this is a conservation of energy problem. When the ball is at the top of its trajectory, when v equals zero, the ball has the highest amount of potential energy and no kinetic energy.

So use the fact that $$V_{final}=0$$ to solve for the highest point it reaches and always keep in mind the direction of the acceleration (g). Potential is highest at top of trajectory, KE is zero there too.