# Homework Help: What is the ball's maximum height?

1. Jan 2, 2012

### Casey314stl

1. The problem statement, all variables and given/known data
A ball is thrown upward.
What is its maximum height? Its initial
vertical speed is 10.8 m/s and the acceleration
of gravity is 9.8 m/s^2
Neglect air resistance.

2. Relevant equations

dfinal=dintial+vfinalΔt+ 1/2averageaccelerationΔt^2 or v^2final=v^2initial +2averageaccelerationΔd

3. The attempt at a solution
14.2463

Last edited: Jan 2, 2012
2. Jan 2, 2012

### mtayab1994

No remember that since the object is travelling up it will have a negative speed therefore:

Vf^2=u^2+2is

3. Jan 2, 2012

### Casey314stl

what does u and is stand for?

4. Jan 2, 2012

### mtayab1994

u and v are vectors v is the final speed which in your case will be 0 because the kinetic energy will be 0 at the maximum height so you will be left with 0=speed of the ball -2(speed of gravity)s.

5. Jan 2, 2012

### Casey314stl

so how do i use this to find the height?

6. Jan 2, 2012

### mtayab1994

0=(initial speed)^2-2(speed of gravity)s then once you have this down just solve the equation.

Last edited: Jan 2, 2012
7. Jan 2, 2012

### Casey314stl

I am solving for Vf^2 which is vertical speed final?
so I plug in 0-2(10.8)=-21.6?

8. Jan 2, 2012

### Staff: Mentor

Your second equation looks promising. In more compact terms:

vf2 = vi2 + 2*a*d

where vi is the initial velocity, vf the final velocity, a is the acceleration and d is the distance. The distance d is what you're looking for. What values will you assign to the other variables?