A ball is thrown from a point 1.0 m above the ground. The initial velocity is 19.6 m/s at an angle of 30 degrees above the horizontal. Find the maximum height of the ball above the ground. Calculate the speed of the ball at the highest point in the trajectory.

Answer 1

max height is

h = v^2 sin^2(θ)/2g
= 19.6^2 * sin^2(30°)/(2*9.8)
= 4.9m

the speed at max height is just the horizontal component, so

19.6cos(30°) = 16.97m/s

Answer 2

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