Work and Energy HELP?

Question

A 500 g stone is dropped from the top of a building 120 above the ground.Determine

a)Velocity when reaches ground

b) Kinetic Energy when it hits ground

Please Help. I don't know what formula to use and its so confusiing. Please help solve this step by step

Answer 1

You're a little confused with units:

I'm going to use the metric system:

For example

building height = h=120 meters

500 grams = 0.5 Kg

g=9.8 m/s^2

Use conservation of energy:

potential energy= m*g*h

kinetic energy= 0.5*m*v^2

mgh=(1/2)*m*v^2

solve for v:

v=sqr(g*h*2)=sqr(9.8*120*2)

= 48.5 m/s

KE=0.5*0.5*(48.5)^2

= 588 joules

check:  mgh= 588 joules

Answer 2

The work done on the stone is going to be -MGH.  

This value will be your kinetic energy.    Remember the conservation of energy and that 1/2MV^2 = K

So, Kinitial = 0.  U intial = MGH

Kfinal = 1/2MV^2 U final = 0

MGH = 1/2MV^2

2GH=V^2

V = √2GH

Answer 3

I think the way this question is asked, it's wanting you to use Kinematics to solve for part a)

using v^2 = u^2 + 2as    ( where u = 0)

        v^2 = 2 * 9.8 * 120

         v  = 48m/s

To solve part b)  you would then use the kinetic Energy formula

Ek = 1/2mv^2