Question

this question by someone by earlier.

consider a .62 kg basetball that is dropped from a height of 1 metee avove the floor:

1.What form or forms of energy does the ball have before it is dropped, if an orgin is set on the ground? Be explicit and give a value for any energy and what form or forms of energy does the ball have right before it makes impact with the ground?

This is the results:

When the ball is above the ground at height of 1m

Potential Energy = m*g*h=0.62*9.8*1 = 6.076 J

Kinetic Energy = 0.5*m*v² = 0.5*0.62*0 = 0 because velocity is 0

When ball is just about to hit ground then all of the potential energy will get converted into kinetic energy due to conservation of energy.

Potential Energy will become zero now because height is 0.

Kinetic Energy will be equa to the potential energy at the highest position i.e.

Kinetic Energy = 6.076 J

2.) Now this is a question I need help on: what does it look like on graph for velocity vs time, kinetic energy vs. time , and potential energy vs time for the ball

Answer 1

as the base ball released from rest which is at a height , the energy possesed by the ball is the gravitational potential energy , and due to the gravitational force the potential energy will be converted in to kinetic energy  

we know that the kinetic energy k.e = 0.5*m*v^2

and the velocity of the ball increases as it reaches the ground , and becomes maximum before hiting the ground

and potentila energy is maximum at time t = 0 s , it s value decreases as time goes on and becomes zero on the ground at later time t .