A major leaguer hits a baseball so that it leaves the bat at a speed of 30.0 m/s and at an angle of 36.9o above the horizontal. Ignore air resistance. (a) At what two times is the baseball at a height of 10.0 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?

Solution 17E (a) onsider the motion of equation 2 s = ut + 1/2 at Here u is vertical component of velocity,t is elapsed time,a is acceleration in vertical direction, s is vertical displacement of ball above the point where it left the base ball. Substitute -g for acceleration and u =0v sin s = v sin t 1/2 g t 0 0 2 10 m = (30 m/s)sin (36.9 ) t 1/2 (9.8 m/s ) t (4.90 m/s)t (18 m/s) t + 10 m = 0 Apply quadratic formula t = 1/9.8 18 ± ( 18) 4(4.90)(10.0) s [ ] t = (1.837 ± 1.154)s Hence the ball at 10 m above the point where it left the bat at t = 0.683 sand 1 2 = 2.99 s. 0 (b).Horizontal component= cos(36.9 ) × 30 m/s = 23.99 m/s. 0 Vertical component= sin(36.9 ) × 30 m/s = 18.01 m/s. 2 2 (c).magnitude of baseball (23.99) + (18.01) = 29.9987 m/s .