A major leaguer hits a baseball so that it leaves the bat at a speed of 33.0 m/s and at an angle of 38.5° above the horizontal. Ignore air resistance.

Question

(a) At what two times is the baseball at a height of 10.7 m above the point at which it left the bat?
(b) Calculate the horizontal and vertical components of the baseball's velocity at each of the two times calculated in part (a).
(c) What are the magnitude and direction of the baseball's velocity when it returns to the level at which it left the bat?

Answer 1

(a) 33 sin38.5° t - 4.9t^2 = 10.7
(b) vertical: 33 sin38.5° - 9.8t
horizontal: 33 cos38.5°
(c) same speed, but negative angle