The given equation for the height of the ball is:

$h = 1 + 15t - 5t^2$

We are tasked with finding the values of $t$ when the ball's height is 6 meters. To solve this, substitute $h = 6$ into the equation:

$6 = 1 + 15t - 5t^2$

Simplify the equation:

$6 = 1 + 15t - 5t^2 \implies 5t^2 - 15t + 5 = 0$

Now, divide the entire equation by 5:

$t^2 - 3t + 1 = 0$

Next, we will solve this quadratic equation using the quadratic formula:

$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For the equation $t^2 - 3t + 1 = 0$, the coefficients are:

• $a = 1$
• $b = -3$
• $c = 1$

Substitute these values into the quadratic formula:

$t = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(1)}}{2(1)}$ $t = \frac{3 \pm \sqrt{9 - 4}}{2}$ $t = \frac{3 \pm \sqrt{5}}{2}$

Thus, the two possible values for $t$ are:

$t = \frac{3 + \sqrt{5}}{2} \quad \text{or} \quad t = \frac{3 - \sqrt{5}}{2}$

Now, compute these values:

$t_1 = \frac{3 + \sqrt{5}}{2} \approx 2.618$ $t_2 = \frac{3 - \sqrt{5}}{2} \approx 0.382$

Therefore, the two values of $t$ when the ball's height is 6 meters are approximately:

$t_1 \approx 2.62 \quad \text{and} \quad t_2 \approx 0.38$

These are the times when the ball is at a height of 6 meters.

Would you like more details or clarifications? Here are some related questions:

1. How does the equation for the height of the ball change if the initial velocity or height changes?
2. What does the vertex of the quadratic function represent in the context of this problem?
3. How can you derive the time when the ball reaches its maximum height?
4. What is the effect of the negative coefficient for $t^2$ on the trajectory of the ball?
5. How do we interpret the two solutions for $t$ in terms of the ball’s motion?

Tip: When dealing with projectile motion, the highest point of the trajectory corresponds to the vertex of the parabola defined by the height function.