Kinetic energy of baseball at highest point of throw

[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. Homework 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?

 

[Redbelly98]

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.

 

[redhot209]

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

 

[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.

 

[redhot209]

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

 

[Redbelly98]

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.

 

[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 mv_x²

 

[tan90]

V_x 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)