To calculate the velocity of the ball when it hits the floor, we can use the equation of motion derived from the law of conservation of mechanical energy. This means the ball's potential energy at the initial height is converted into kinetic energy just before it hits the ground.

Given dataTo calculate the velocity of the ball when it hits the floor, we can use the equation of motion derived from the law of conservation of mechanical energy. This means the ball's potential energy at the initial height is converted into kinetic energy just before it hits the ground.

Given

• Mass of the ball, $m = 0.050 \, \text{kg}$
• Initial height, $h = 0.85 \, \text{m}$
• Acceleration due to gravity, $g = 9.81 \, \text{m/s}^2$

We can ignore air resistance, and use the following energy conservation relation:

$mgh = \frac{1}{2} mv^2$

Where:

• $mgh$ is the potential energy at height $h$,
• $\frac{1}{2} mv^2$ is the kinetic energy just before it hits the floor.

Since the mass $m$ cancels out, we are left with:

$gh = \frac{1}{2} v^2$

Solving for $v$ (velocity):

$v = \sqrt{2gh}$

Substituting the values:

$v = \sqrt{2 \times 9.81 \, \text{m/s}^2 \times 0.85 \, \text{m}}$

Let's calculate this.The velocity of the ball just before it hits the floor is approximately $4.08 \, \text{m/s}$.

Would you like further clarification on any part of this process?

Here are 5 questions to expand on this concept:

1. How does the presence of air resistance affect the final velocity?
2. What would the final velocity be if the ball were released from a greater height, say 2 meters?
3. How is the conservation of energy principle applied in this problem?
4. What is the significance of mass in calculating the final velocity?
5. What would happen if the ball was thrown downwards with an initial velocity?

Tip: When ignoring air resistance, all objects (regardless of mass) fall with the same acceleration due to gravity.