To tackle the problem step-by-step:

Given:

The height of the projectile as a function of time is:

$h(t) = 2 + 21.5t - 4.9t^2$

Where:

• $t$ is time in seconds,
• $h(t)$ is height in meters.

(c) Find the Maximum Height:

The maximum height occurs at the time when the projectile reaches its peak, i.e., at the time when the velocity is 0. The velocity $v(t)$ is given by the derivative of the height function:

$v(t) = \frac{dh(t)}{dt} = 21.5 - 9.8t$

From part (b), the time at maximum height is $t = 2.19$ seconds. So, we substitute this into the height equation $h(t)$:

$h(2.19) = 2 + 21.5(2.19) - 4.9(2.19)^2$

Let’s calculate it:

$h(2.19) = 2 + 21.5(2.19) - 4.9(4.7961)$ $h(2.19) = 2 + 47.785 - 23.5029$ $h(2.19) \approx 26.28 \, \text{m}$

So, the maximum height is approximately 26.28 meters, not 25.59 m as previously calculated.

(d) When Does the Projectile Hit the Ground?

To find when the projectile hits the ground, set $h(t) = 0$ and solve for $t$:

$0 = 2 + 21.5t - 4.9t^2$

This is a quadratic equation. Rearranging:

$4.9t^2 - 21.5t - 2 = 0$

Now, solve this quadratic equation using the quadratic formula:

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

Where $a = 4.9$, $b = -21.5$, and $c = -2$:

$t = \frac{-(-21.5) \pm \sqrt{(-21.5)^2 - 4(4.9)(-2)}}{2(4.9)}$ $t = \frac{21.5 \pm \sqrt{462.25 + 39.2}}{9.8}$ $t = \frac{21.5 \pm \sqrt{501.45}}{9.8}$ $t = \frac{21.5 \pm 22.39}{9.8}$

So, we have two possible solutions:

$t = \frac{21.5 + 22.39}{9.8} \quad \text{or} \quad t = \frac{21.5 - 22.39}{9.8}$ $t = \frac{43.89}{9.8} \quad \text{or} \quad t = \frac{-0.89}{9.8}$ $t \approx 4.48 \quad \text{or} \quad t \approx -0.09$

Since negative time doesn’t make sense in this context, the projectile hits the ground after approximately 4.48 seconds.

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Would you like me to explain any steps in more detail? Here are 5 related questions for further practice:

1. What is the velocity of the projectile at its maximum height?
2. If the initial velocity were increased, how would that affect the maximum height and time to hit the ground?
3. How would the equation change if the projectile were launched from 10 meters above ground instead of 2 meters?
4. Can you derive the formula for velocity from the height function?
5. What are the units of each term in the height equation $h(t) = 2 + 21.5t - 4.9t^2$?

Tip: When solving quadratic equations, always check the physical relevance of your solutions, especially in time-dependent problems like this one!