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)
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.
1 answer
1 answer