The height of the ball is represented by the equation d = 1/2at^2

The veritcal component of the velocity of the ball from it's maximum hieght to the ground is represented by the equation

vf^2 = vi^2 + 2ad.

The velocity of the ball is repsented by Vr sin 30 = vf

So now we make three substitutions (both d's are the same, max height)

t = 1.5 seconds (time to the max hieght of ball a)

so we get...

vf = ((2a(1/2at^2))^.5)/sin 30

This gives a velocity of 29.4 m/s.

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