Find the second and third values (after the minimal one) of velocity too. The ball is kicked horizontally, near to the Earth's surface.

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I haven't been back in awhile and I saw this was not solved so I decided to solve it. It really looks a lot harder than it is.

x-x0=1/2at^2 (there is no initial velocity in the y direction)

-1 meter=1/2(-9.8)(m/s^2)t^2

1 meter=1/2(9.8)(m/s^2)t^2

t^2=1/4.9 s^2

t=(1/4.9)^1/2 s=0.45 s

The horizontal distance traveled is 2(.5m)=1 m

So the initial velocity is 1m/0.45 s=2.21 m/s

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Well done! When we understand the principle of independence of motion, this is an easy task indeed. Thus, this is the minimal velocity. The second minimal velocity we obtain when the ball once hit the right wall and twice the left one...

Hint: The ball is free falling independently on its motion to the left. Thus, knowing the height h (together with gravitational acceleration g) is enough to calculate the time ball needs to reach the corner. Than we can use the same fact to calculate initial velocity...

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Seriously. Have you enhanced the original Lysergic Acid Diethylamide formula?

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