In: Physics
A major leaguer hits a baseball so that it leaves the bat at a speed of 29.0 m/s and at an angle of 36.5 degrees above the horizontal. You can ignore air resistance. Calculate the vertical component of the baseball's velocity at two times. t1 = .732s and t2= 2.79s. These are the two times when the baseball was at a height of 10.0 mm above the point at which it left the bat. There should be two answers in m/s.
Solution
At t1, 10.076m/s (It is upward)
At T2, -10.092m/s (Negative sign indicates it is directed downward. If magnitude is only needed take its magnitude as 10.092m/s)