How Is Kinetic Energy Calculated at the Peak of a Baseball's Trajectory?

[kiwikahuna]

Homework Statement

an outfielder throws a 2.15 kg baseball at a speed of 113 m/s and an initial angle of 11.2 degrees. What is the kinetic energy of the ball at the highest point of its motion.

Homework Equations

KE = 1/2mv^2

KE(initial) + PE(initial) = KE (final) + PE (final)

The Attempt at a Solution

KE = 1/2 (2.15 kg) (113* sin 11.2)^2
KE =517.866

My method seems too simple. Should I be using the conservation of energy equation and if so, how should I go about doing it?

 

[Archduke]

Almost right - have a look at you sine/cosine.

You know that you can consider the velocity components in the horizontal and vertical directions independently. What are the forces acting along each direction of the velocity components? When the ball is at its highest point, what can you say about the ball's vertical velocity? And what about the horizontal velocity?

As I say, you've almost got it right, just look at splitting the velocity into components again.

 

[kiwikahuna]

when the ball is at its highest point, the ball's vertical velocity would be zero? Should I have used: KE = 1/2 (2.15 kg) (113* cos 11.2)^2 instead?

 

[Doc Al]

That is correct.