A 5.1 oz baseball is hit from 2 ft above the ground. It has an initial velocity of 165 ft/s at an angle 40 degrees above horizontal. Find the Kinetic Energy of the ball at its maximum height.
Work:
At max height, the vertical velocity is zero. the horizontal velocity is constant since there is no horizontal acceleration [V=165*cos(40)]. Kinetic Energy (KE)=(1/2)mv^2 {(1/2)(5.1/16)(165*cos(40))}=2546.2 lb ft^2/s^2
I was told that is not the correct answer.