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  • 11-02-2021
  • Physics
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If the kinetic energy of the 40kg box is 784 J, what is the velocity before it strikes the ground?

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GeorgePP
GeorgePP GeorgePP
  • 11-02-2021

Answer:

Explanation:

[tex]KE=\frac{1}{2}mv^2[/tex]

[tex]784=\frac{1}{2}(40)v^2[/tex]

[tex]784=20v^2[/tex]

[tex]39.2=v^2[/tex]

[tex]v=6.26m/s[/tex]

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