posted by Sherry on .
A ball is thrown from a point 1.1 m above the ground. The initial velocity is 20.0 m/s at an angle of 32.0° above the horizontal.
(a) Find the maximum height of the ball above the ground.
(b) Calculate the speed of the ball at the highest point in the trajectory.
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)