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g A projectile of mass 3 kg is launched horizontally from an initial height 3 m with an initial velocity 10 m/s. This velocity in the x direction is preserved when you ignore air resistance. The projectile still accelerates in the vertical y direction toward the ground, but this is exactly the energy lost from potential energy. Energy is conserved as long as you use the total mechanical energy equation. What is the final kinetic energy as the projectile just reaches the ground

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Cricetus

Answer:

The kinetic energy at ground will be "238.2 J".

Explanation:

The given values are:

mass,

m = 3 kg

Initial height,

h = 3 m

Initial velocity,

v = 10 m/s

By using the conservation of energy at points A and B,

⇒  [tex]E_A=E_B[/tex]

⇒  [tex]mgh+\frac{1}{2}mv^2=k_B[/tex]

On substituting the values, we get

⇒  [tex]3\times 9.8\times 3+\frac{1}{2}\times 3\times (10)^2=k_B[/tex]

⇒             [tex]88.2+0.5\times 3\times 100=k_B[/tex]

⇒                            [tex]88.2+150=k_B[/tex]

⇒                                    [tex]238.2 =k_B[/tex]

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