Question

A ballplayer standing at home plate hits a baseball that is caught by another player at the same height above the ground from which it was hit. The ball is hit with an initial velocity of 32.1 m/s at an angle of 60.0° above the horizontal. How high will the ball rise than where it was hit?

Answer 1

Answer:

Maximum height reached = 39.39 m

Explanation:

At maximum height

We have equation of motion v² = u² + 2as

Consider the vertical motion of baseball,

Initial velocity, u = 32.1sin60=27.8 m/s

Final velocity, v = 0 m/s

Acceleration, a = -9.81 m/s²

Substituting

             v² = u² + 2as  

             0 = 27.8² + 2 x (-9.81) x s

             s = 39.39 m

Maximum height reached = 39.39 m