Two 100 kilogram balls are mounted on a seesaw and this seesaw is in balanced position .Now I want to know that how much force I should apply on the right side arm of this balanced seesaw so that the left side 100 kilogram weight ball could jump in the air up to 5 centimeter.

This question is regarding a project and it is not a homework related.

I assume what you mean is you want the 100 Kg ball on the left to fly up off the seesaw when the right side hits the ground and stops - correct? You'll need to tell us the travel distance over which the force on the right side operates - in other words how far does the right side travel from the startng equilibrium point before hitting the ground?

The travel distance of right side will be 10 centimeter from the ground.

You will need about 1000 N force applied downward on the right hand side of the seesaw.

Dear sir,can you please provide me some equations and formulas regarding your reply.

Use energy principles:

1. The velocity you want at the instant the seesaw stops is v^2 = 2gh, where h = 0.05 meters (the height the 100 Kg mass needs to rise above the stopping point).

2. The work put into the system is F times d, where F is the applied force and d is the distance the force is applied over, which is 0.1m.

3. Conservation of energy:
Change in kinetic energy of the two masses + change in potential energy of the two masses = work applied.

Can you take it from here?