Posted by Sherry on Thursday, September 13, 2012 at 11:53am.
the vertical component of the velocity (Vv)is __ 20.0 * sin(32.0º)
the horizontal component (Vh) is __ 20.0 * cos(32.0º)
the time (T) to max height is __ Vv / g
a) Hmax = (-.5 g T^2) + (Vv * T) + 1.1
b) at Hmax, the ball has no vertical velocity; only horizontal (Vh)
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