3. A baseball (m = 0.150 kg) is thrown with an initial speed of 30.0 m/s at an angle of 37.0 degrees from the horizontal at y = 0 m. (a) Find the initial KE of the ball. (b) Use conservation of energy (not kinematic equations) to find the maximum height reached by the ball. (Hint: Does the ball have any velocity at its maximum height?)
KEf + Uf = KEi + Ui
(.5)mvf² + mgyf = (.5)mvi² + mgyi
The Attempt at a Solution
(a) Part a I got the correct answer, but can you tell me if I did it the correct way. I did (.5)(0.15)(30)² = 67.5J
(b) What I tried was using KEf + Uf = KEi + Ui . For the velocity at the top I accounted for the angle and got that the velocity in the x direction was 24.0m/s So I did:
(.5)(0.15)(24.0)² + (.5)(9.8)y = 67.5 (from part a) + (0.15)(9.8)(0)
Solving for y I get 4.96m, but that is not the correct answer. I'm not sure if I am doing it the wrong way or if I made some error somewhere, so any help would be appreciated.