A ball is thrown straight upward and returns to the thrower's hand after 2.20 s in the air. A second ball is thrown at an angle of 31.0° with the horizontal. At what speed must the second ball be thrown so that it reaches the same height as the one thrown vertically?

Answer 1

You need the vertical velocities the same.

time in air for the second ball.
vertical velocity at the top is zero.

Vf=Vsin31-g*t
Vf=0, so t=Vsin31/g but this time is 1.1 seconds, so solve for V to make that happen.

Answer 2

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