To find the initial velocity of the ball, we can solve this problem by analyzing it in two parts: horizontal and vertical motion.

Step 1: Analyze vertical motion

The ball falls a vertical distance of 0.912 meters under the influence of gravity. The vertical motion can be described by the equation:

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

Where:

• $y$ is the vertical distance (0.912 meters),
• $g$ is the acceleration due to gravity (approximately $9.81 \, \text{m/s}^2$),
• $t$ is the time of flight.

We can solve this equation for $t$, the time it takes for the ball to fall:

$0.912 = \frac{1}{2} \times 9.81 \times t^2$

$t^2 = \frac{2 \times 0.912}{9.81}$

$t^2 = 0.1859$

$t = \sqrt{0.1859} = 0.431 \, \text{seconds}$

Step 2: Analyze horizontal motion

The horizontal motion is uniform (constant velocity) since no forces are acting on the ball horizontally after it leaves the desk.

The horizontal distance $x$ traveled by the ball is given by:

$x = v_0 \times t$

Where:

• $x$ is the horizontal distance (1.31 meters),
• $v_0$ is the initial horizontal velocity (what we need to find),
• $t$ is the time of flight (0.431 seconds).

Rearranging to solve for $v_0$:

$v_0 = \frac{x}{t} = \frac{1.31}{0.431} = 3.04 \, \text{m/s}$

Final Answer:

The initial velocity of the ball is approximately 3.04 meters per second.

Would you like more details on any part of this solution or a different problem analyzed?

Here are 5 related questions:

1. How does the vertical distance affect the time of flight in projectile motion?
2. What would happen to the horizontal distance if the initial velocity doubled?
3. Can we solve this problem if the horizontal velocity is not constant?
4. How would air resistance impact the horizontal distance traveled?
5. What is the relationship between time of flight and vertical displacement?

Tip: In projectile motion, vertical and horizontal components are treated independently, with gravity only influencing vertical motion.