Question

If you drop a 0.05 kg egg from 2m, what do you predict would be the velocity of the egg before it hits the ground, neglecting air resistance?

(I'm a bit confused, can someone explain?)​

Answer 1

This kind of problems becomes extremely easy if you use the consevation of energy. We know that, at any given moment, the total energy is constant.

Specifically, the energy is given by the sum of the kinetic and potential energy:

`E=K+U=(1)/(2)mv^2+mgh`

At the beginning, the egg has no velocity (
`v=0`), it's mass is 0.05, and its height is 2m, so the initial energy is

`E_i=(1)/(2)\cdot 0.05\cdot 0^2+0.05\cdot g\cdot 2 = 0.1g`

When the egg hits the ground, its height has become zero. So, the total energy is

`E_f=(1)/(2)\cdot 0.05\cdot v^2+0.05\cdot g\cdot 0 = 0.1v^2`

The initial and final energies must be the same, so we have

`0.1v^2=0.1g\iff v^2=g \iff v=\pm√(g)`

Discarding the negative solution, which woudln't make sense, we have

`v=√(g)`

where
`g` is the acceleration due to gravity

Answer 2

Step-by-step explanation:

i think you have to find the velocity by using its wight times the gravity of the earth times the distance it fell?

look in google for a formula the use these sign, w and g ,d