A physics book of unknown mass is dropped 5.00 m. What speed does the book have just before it hits the ground? Assume that air resistance is negligible.

To solve this problem, we can use the physics principle of conservation of energy. Specifically, we'll use the formula for gravitational potential energy and kinetic energy.

1. First, we calculate the initial potential energy of the book. The formula for gravitational potential energy is PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height. In this case, we don't know the mass of the book, but we know that gravity is approximately 9.8 m/s² and the height is 5.00 m. So the potential energy is m * 9.8 * 5.00.

2. When the book is falling, it's converting its potential energy into kinetic energy. Just before it hits the ground, all its potential energy has been converted into kinetic energy. The formula for kinetic energy is KE = 1/2 mv², where m is the mass and v is the velocity.

3. Since the potential energy equals the kinetic energy, we can set the two equal to each other and solve for v (the velocity).

So, m * 9.8 * 5.00 = 1/2 m * v².

You'll notice that the mass (m) is on both sides of the equation, so it cancels out. This leaves us with:

9.8 * 5.00 = 1/2 * v².

4. To solve for v, we first multiply both sides by 2 to get rid of the 1/2 on the right side:

2 * 9.8 * 5.00 = v².

5. Finally, we take the square root of both sides to solve for v:

v = sqrt(2 * 9.8 * 5.00).

If you do this calculation, you'll find that the velocity of the book just before it hits the ground is approximately 10 m/s.