physics
posted by jennifer on .
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?

You need the vertical velocities the same.
time in air for the second ball.
vertical velocity at the top is zero.
Vf=Vsin31g*t
Vf=0, so t=Vsin31/g but this time is 1.1 seconds, so solve for V to make that happen. 
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