To find the value(s) of $t$ when the cannonball is on the ground, i.e., when $y = 0$, we need to solve the quadratic equation:

$y = 64t - 8t^2$

Set $y = 0$:

$0 = 64t - 8t^2$

This is a quadratic equation in the form of $0 = at^2 + bt + c$. Let's simplify and solve it step by step.

Step 1: Factor the equation

First, factor out the common term, which is $-8t$:

$0 = -8t(t - 8)$

Step 2: Solve for $t$

Now, solve for $t$ by setting each factor equal to zero:

$-8t = 0 \quad \text{or} \quad (t - 8) = 0$

From $-8t = 0$:

$t = 0$

From $t - 8 = 0$:

$t = 8$

Conclusion:

The values of $t$ when the cannonball is on the ground (i.e., $y = 0$) are $t = 0$ and $t = 8$.

Interpretation:

• At $t = 0$, the cannonball was just launched from the ground.
• At $t = 8$, the cannonball returns to the ground after reaching its maximum height.

Would you like more details or clarifications on any step?

Here are 5 related questions:

1. What is the maximum height the cannonball reaches, and at what time?
2. How can we determine the total time the cannonball is in the air?
3. How does changing the coefficient of $t^2$ affect the flight of the cannonball?
4. Can you graph this quadratic equation and describe its shape?
5. How does this scenario relate to real-life projectile motion?

Tip: For quadratic equations, factoring is a quick method when the equation can be easily factored, but for more complex cases, the quadratic formula is a powerful alternative.