Let's analyze the given problem systematically. We are provided with the height equation of the can after it was thrown off a building, given by:

$H(t) = (t + 2)(5 - t)$

where $H(t)$ represents the height of the can in terms of time $t$ in seconds.

1. Sketch the Graph:

To sketch the graph of $H(t) = (t + 2)(5 - t)$, we can first expand and rewrite the equation:

$H(t) = t(5 - t) + 2(5 - t) = 5t - t^2 + 10 - 2t = -t^2 + 3t + 10$

This is a quadratic equation of the form $H(t) = -t^2 + 3t + 10$. Since the coefficient of $t^2$ is negative, the parabola opens downward.

2. When does the can land on the road?

The can lands on the road when its height is zero ($H(t) = 0$). We solve for $t$ by setting the equation to zero:

$(t + 2)(5 - t) = 0$

This gives us two solutions:

1. $t + 2 = 0$ ⟹ $t = -2$
2. $5 - t = 0$ ⟹ $t = 5$

Since negative time doesn’t make sense in this context, the can lands on the road when $t = 5$ seconds.

3. After how many seconds is the can again at the same height as the roof?

The height of the can at the moment it was thrown (initial height) is given by:

$H(0) = (0 + 2)(5 - 0) = 2 \cdot 5 = 10$

We want to find when $H(t) = 10$ again:

$-t^2 + 3t + 10 = 10$

$-t^2 + 3t = 0$

$t(-t + 3) = 0$

The solutions are:

1. $t = 0$ (initial moment)
2. $-t + 3 = 0$ ⟹ $t = 3$

So, the can is again at the same height as the roof after $t = 3$ seconds.

4. Over which set of t-values is the graph an increasing function?

To find where the graph is increasing, we take the derivative of $H(t)$ and set it to greater than zero:

$H'(t) = -2t + 3$

Set $H'(t) > 0$:

$-2t + 3 > 0$

$-2t > -3$

$t < \frac{3}{2}$

So, the function is increasing for $0 \leq t < \frac{3}{2}$.

5. What is the maximum height reached by the can?

The maximum height of a downward-opening parabola is at its vertex. The vertex formula for $H(t) = at^2 + bt + c$ is:

$t = -\frac{b}{2a}$

For our equation $H(t) = -t^2 + 3t + 10$, $a = -1$ and $b = 3$:

$t = -\frac{3}{2(-1)} = \frac{3}{2}$

The maximum height is:

$H\left(\frac{3}{2}\right) = -\left(\frac{3}{2}\right)^2 + 3\left(\frac{3}{2}\right) + 10$

$H\left(\frac{3}{2}\right) = -\frac{9}{4} + \frac{9}{2} + 10$

$H\left(\frac{3}{2}\right) = \frac{-9 + 18 + 40}{4}$

$H\left(\frac{3}{2}\right) = \frac{49}{4} = 12.25$

The maximum height reached by the can is 12.25 meters.

Summary of Answers:

1. The can lands on the road at $t = 5$ seconds.
2. The can is again at the same height as the roof after $t = 3$ seconds.
3. The graph is increasing for $0 \leq t < \frac{3}{2}$.
4. The maximum height reached by the can is 12.25 meters.

Would you like further details or have any more questions?

Relative Questions:

1. How would the graph change if the can were thrown with an initial upward velocity?
2. What would happen if the can were thrown from a different initial height?
3. How does air resistance affect the trajectory of the can?
4. What if the equation included a horizontal component of the motion?
5. Can we determine the initial speed of the can from the given equation?

Tip:

Always check the signs of the coefficients in a quadratic equation to understand the direction (upward or downward) of the parabola's opening.