case of rebounding of ball....

In a frictionless world, a ball dropped from a height of 5 m would rebound 5 m. However, air resistance (friction encountered while traveling through the atmosphere) causes enough energy loss in proportion to distance traveled to make the ball rebound 2 m less. 2 m lost over 8 m traveled = ¼ or 0.25 or 25% of the energy being lost to friction.

According to the question, you want the ball to travel 5 m down and 5 m up for a total of 10 m, by adding enough velocity to compensate for the expected 25% energy loss. 25% of 10 m is 2.5 m. Hold on to this fact because you'll need to use it later.

Now you've got potential energy on one hand (PE = mgh) and kinetic energy on the other (KE = ½mv²). You want to find your starting velocity (v) that creates the same amount of kinetic energy as the potential energy lost to air resistance. (For the purposes of this problem we are going to assume that the conversion from kinetic to potential energy is lossless. It isn't perfectly efficient in real life.)

PE = KE

mgh = ½mv²

gh = ½v²

2gh = v²

v = √(2gh)

After all of that conceptual work, the calculation that you need is:

What is v when the Earth's gravitational constant g = 9.8 m/s² (approximately) and height h = 2.5 m?

v = √[2(9.8 m/s²)(2.5 m)] = √[(5 m)(9.8 m/s²)] = √(49 m²/s²) = 7 m/s

So from a height of 5 meters, you'd need to throw straight down at 7 meters per second if you want the ball to bounce back exactly 5 meters.

Now, can you answer a similar question? The air on Mt. Everest is so thin that there's less air resistance. Suppose a ball on Mt. Everest is dropped from a height of 20 meters. It rebounds to 16 meters. With what velocity must the ball on Mt. Everest be thrown from 50 meters high, so that it rebounds to its original height?

See also - Khan Academy: Elastic and inelastic collisions