We are given the following data from the image:

• Horizontal distance $x = 560 \, \text{m}$
• Gravitational acceleration $g = 10 \, \text{m/s}^2$
• Time of flight $t = 8 \, \text{s}$

Assuming the projectile was fired horizontally (since it follows a parabolic trajectory), we can calculate the initial velocity using the equation for horizontal motion:

$x = v_{0x} \cdot t$

Where:

• $x$ is the horizontal distance,
• $v_{0x}$ is the horizontal component of the initial velocity,
• $t$ is the time.

Step 1: Calculate the initial velocity in the horizontal direction

We rearrange the equation to solve for $v_{0x}$:

$v_{0x} = \frac{x}{t}$

Substitute the given values:

$v_{0x} = \frac{560 \, \text{m}}{8 \, \text{s}} = 70 \, \text{m/s}$

Thus, the initial horizontal velocity of the projectile is $v_{0x} = 70 \, \text{m/s}$.

Would you like further details on this calculation or any clarifications? Here are some related questions:

1. How does the time of flight affect the horizontal velocity?
2. What would the trajectory look like if the projectile was launched at an angle?
3. Can you determine the maximum height reached by the projectile in this case?
4. How would increasing the initial velocity affect the range of the projectile?
5. What is the vertical motion equation for this type of problem?

Tip: In projectile motion, horizontal and vertical motions are independent of each other. Horizontal velocity remains constant when air resistance is neglected.