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Awhat is the kinetic energy of a 1 kg ball thrown in the air with an initial velocity of 30 m/s
BHow high did the ball travel? KE=PE

Verified Answer :

By Definition of Kinetic Energy, it is valid:

\(E_c= \frac{mv^2}{2} \\ E_c= \frac{1\times 30^2}{2} \\ \boxed {E_c=450~J}\)

     Using the Torricelli's Equation in vertical direction on the maximum height, remember that the final velocity at this point is zero, we have:

\(v^2=v_{o}^2+2a\Delta S \\ 0=30^2+2(-10)H \\ H= \frac{900}{20} \\ \boxed {H=45~m}\)

If you notice any mistake in my english, please let me know, because my english level is still low.

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