# Help with 1 problem

1. Oct 10, 2008

### redhot209

1. An outfield throws a 1.51 kg baseball at a speed of 118 m/s and an initial angle of 41.2 degrees. What is the kinetic energy of the ball at the highest point of its motion? Answer in units of J.

2. Relevant equations:None

3. The attempt at a solution: I used the equation K=1/2mv^2
1/2(1.51)(118)^2= 10,512.62 J but answer was wrong. Any ideas what I did wrong?

2. Oct 10, 2008

### Redbelly98

Staff Emeritus
You calculated the initial kinetic energy. They are asking for the kinetic energy when the ball is at the height of it's trajectory. v will be different than 118 m/s there.

3. Oct 10, 2008

### redhot209

So how would I find V, as the final velocity?

4. Oct 10, 2008

### Rake-MC

Well at it's highest point in it's trajectory, it's not rising nor falling, am I right?

Therefore the y component of velocity = 0.

Take it from there.

5. Oct 11, 2008

### redhot209

sorry to inform you but the velocity isn't zero.

6. Oct 11, 2008

### Redbelly98

Staff Emeritus
Rake-MC is correct that the y component of velocity is zero at the highest point.

redhot209 is also correct, the velocity is not zero.

Both statements are correct, since velocity and y-component of velocity are not the same thing.

7. Oct 11, 2008

### tan90

y component of the velocity is zero and you still need to calculate the x-component, which you can use to calculate K.E. by using the formula, K.E. = 1/2 mvx2

8. Oct 11, 2008

### tan90

Vx always remains the same in a parabolic trajectory, so it will be same as initial velocity along x direction which will be 118 cos(41.2)