hysics book, mass unknown, is dropped 4.50 m . What speed does the book have just before it hi ground?

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1. The principle of conservation of energy states that the gravitational potential energy at the start is converted entirely into kinetic energy just before impact. This gives:
   $$
   mgh = \frac{1}{2}mv^2
   $$

2. Canceling the mass $$ m $$ from both sides (since it is unknown and non-zero):
   $$
   gh = \frac{1}{2}v^2
   $$

3. Solve for $$ v $$:
   $$
   v = \sqrt{2gh}
   $$

4. Substitute $$ g = 9.8 \, \text{m/s}^2 $$ and $$ h = 4.50 \, \text{m} $$:
   $$
   v = \sqrt{2 \times 9.8 \, \text{m/s}^2 \times 4.50 \, \text{m}} = \sqrt{88.2 \, \text{m}^2/\text{s}^2}
   $$

5. Calculate the square root:
   $$
   v \approx 9.39 \, \text{m/s}
   $$

Thus, the speed of the book just before it hits the ground is approximately $$ 9.39 \, \text{m/s} $$.
The book hits the ground at about 9.39 meters per second.

hysics book, mass unknown, is dropped 4.50 m . What speed does the book have just before it hi ground?

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