Free help with homework

Why join Brainly?

  • ask questions about your assignment
  • get answers with explanations
  • find similar questions



This Is a Certified Answer

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
Let the stone of mass m be at a height h and be dropped from rest at t = 0 sec.

Initial potential energy = PE1 = m g h
initial kinetic energy = 1/2 m 0^2 = 0

Let the stone drop by a distance s in time  t.  As per the laws of kinetics we have
       2 g s = (v^2 - u^2 ) = v^2

Potential energy at time t and height of (h - s) is   PE2 = m g (h - s)
Kinetic energy = KE2 = 1/2 m v^2 = 1/2 m (2 g s) = m g s

Total energy = m g (h - s) + m g s = m g  h

All along the path ,  from t = o to  t = root(2h/g) ,  and height h to 0, the total energy remains as  m g h.

The other forms of energy in the object are assumed to be constant and not varying.

2 3 2
please click on thank you link and select best answer
The Brain
  • The Brain
  • Helper
Not sure about the answer?
Learn more with Brainly!
Having trouble with your homework?
Get free help!
  • 80% of questions are answered in under 10 minutes
  • Answers come with explanations, so that you can learn
  • Answer quality is ensured by our experts