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A 0.5 kg ball is thrown upwards with an initial velocity of 9 m/s. How high will the ball reach before it starts falling again?

Respuesta :

Answer:

About 4.13 m

Explanation:

[tex]v^2=v_0^2+2a\Delta x[/tex]

[tex]\Delta x=\frac{v_0^2}{2g}[/tex]

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