To determine the initial velocity of the ball as it rolls horizontally off the table, we can solve the problem using the following physics principles:

1. Vertical Motion: The ball falls a vertical distance $h = 0.750$ meters due to gravity. Using the equation for free fall:

$h = \frac{1}{2} g t^2$

Where:

• $h = 0.750$ m
• $g = 9.8 \, \text{m/s}^2$ (acceleration due to gravity)
• $t$ is the time it takes to hit the ground

Solving for $t$:

$t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 0.750}{9.8}} \approx 0.39 \, \text{seconds}$

2. Horizontal Motion: The horizontal distance the ball travels is $d = 0.470$ meters, and the horizontal velocity $v_0$ is constant since there’s no horizontal acceleration. The relation between horizontal distance and velocity is:

$d = v_0 t$

Solving for $v_0$:

$v_0 = \frac{d}{t} = \frac{0.470}{0.39} \approx 1.21 \, \text{m/s}$

So, the initial velocity of the ball is approximately 1.21 meters per second.

Would you like more details on any part of this explanation?

Here are 5 related questions to explore further:

1. What would happen to the initial velocity if the table height were doubled?
2. How would air resistance affect the ball's motion?
3. How would the horizontal distance change if the initial velocity were doubled?
4. How do you calculate the horizontal distance for a projectile launched at an angle?
5. What role does gravity play in horizontal projectile motion?

Tip: In projectile motion, horizontal and vertical components of motion are treated independently.