# At what speed must the second ball be thrown so that it reaches the same height as the one thrown vertically? Answer in units of m/s. A ball is thrown straight upward and returns to the throwers hand after 3 s in the air. A second ball is thrown at an angle of 30 degrees with the horizontal. The acceleration of gravity is 9.8 m/s^2. 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

The velocity of the ball is repsented by Vr sin 30...

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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

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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