A pitcher throws a ball at 44.2 m/s horizontally. How much drop time until the catcher gets it 15m away?

Answer 1

The position of the ball at any time t is
x = (Vocosx) t and y= -4.9t^2 +(Vo sin x) t, where Vo = initial velocity, x = angle ball thrown at = 0, and t = time in seconds.

So 15 = 44.2cos(0) t
t = 15/44.2 = .3393 seconds
At .3393 seconds, y = -4.9(.3393)^2 = .56m