EXPLAIN

A rubber ball (mass 0.30 kg) is dropped from a height of 1.2 m onto the floor. Just after bouncing from the floor, the ball has a speed of 3.6 m/s.
(a) What is the magnitude and direction of the impulse imparted by the floor to the ball?

so to find v1=square root of 2gh
V1*m +V2*m=2.5

WHERE did the square root of 2gh come from?

Remember the equation

Vf^2=Vi^2+2gh ? If Vi is zero, then...Vf=

Where did it come from..Conservation of energy.
The final KE at impact= the intial PE
1/2 mv^2=mgh

The equation to find the initial velocity of the ball before it bounces can be derived from the principle of conservation of energy. The potential energy of an object at a certain height is given by the formula mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height.

When the ball is dropped from a height, it converts its potential energy into kinetic energy as it falls towards the ground. At the instant just before bouncing, all of the potential energy is converted into kinetic energy. According to the conservation of energy, the potential energy at the top of its fall is equal to the kinetic energy just before it hits the ground.

The formula for the initial velocity of an object can be derived from the equation for kinetic energy. The kinetic energy of an object is given by the formula (1/2)mv^2, where m is the mass of the object and v is the velocity.

Setting the potential energy equal to the kinetic energy, we have:

mgh = (1/2)mv^2

Canceling out the mass (m) on both sides, we get:

gh = (1/2)v^2

Simplifying further, we can solve for v:

v^2 = 2gh

And taking the square root of both sides, we obtain:

v = √(2gh)

This is where the square root of 2gh comes from in the equation you mentioned. It allows us to calculate the initial velocity of the ball just before it hits the ground based on its height.

The equation v1 = sqrt(2gh) comes from the principle of conservation of mechanical energy. It states that the initial mechanical energy of the ball (when it is at a height h) is equal to the sum of its kinetic energy just after bouncing (which is equal to 1/2 * m * v1^2) and its potential energy just before bouncing (which is equal to m * g * h).

In this case, the ball is dropped from a height of 1.2 m, so its initial potential energy is m * g * h = 0.30 kg * 9.8 m/s^2 * 1.2 m = 3.528 J. The final kinetic energy just after the bounce is equal to 1/2 * m * v1^2, and since the speed is given as 3.6 m/s, we can solve for v1 by using the equation:

1/2 * m * v1^2 = 3.6 m/s * 0.30 kg

Solving this equation for v1 gives us:

v1 = sqrt((2 * 3.6 m/s * 0.30 kg) / (0.30 kg)) = sqrt(2 * 3.6 m/s) = 2.68 m/s

Therefore, the square root of 2gh is used to find the initial velocity (v1) of the ball just before it hits the floor.