A2.80 kg mass is dropped from a height of 4 50 m. Find its kinetic energy (KE) when it is 3.00 m above the ground

Answer 1

I'm going to assume the height is missing a decimal place and it is supposed to say 4.50 m.

The energy that the ball had instantaneously prior to being dropped was Potential Energy (E_P) only and gravitational potential energy is calculated as E_p = mgh = (2.80)(9.81)(4.50) = 123.606 Joules.

Because no outside energy was added and no work was done, the energy instantaneously prior to being dropped is equal to the energy when the mass is 3 m above the ground only at that point, the energy consists of two parts, gravitational Potential Energy (EP) and Kinetic Energy (EK). So, considering the point just prior to the drop as point 1 and the point 3 m above the ground as point 2, we can say:

EP₁ = EP₂ + EK₂

so EK₂ = EP₁ - EP₂

and EP₂ = (2.80)(9.81)(3.00) = 82.404

So EK₂ = 123.606 - 82.404 = 41.2 Joules

Answer 2

KE = mg(H - h) = 2.80kg·9.81m/s²(450m - 3m) = 12278.2 J