Let's go through each question systematically.

Given: $h(t) = 2 + 21.5t - 4.9t^2$ where $h(t)$ is the height in meters after $t$ seconds.

To analyze this, we’ll first find the velocity function by differentiating $h(t)$ with respect to $t$, since velocity is the derivative of the height function with respect to time.

1. Velocity function, $v(t)$: $v(t) = \frac{dh}{dt} = 21.5 - 9.8t$

Part (a) Find the velocity at $t = 2$ and $t = 4$

1. At $t = 2$: $v(2) = 21.5 - 9.8 \times 2 = 21.5 - 19.6 = 1.9 \, \text{m/s}$

2. At $t = 4$: $v(4) = 21.5 - 9.8 \times 4 = 21.5 - 39.2 = -17.7 \, \text{m/s}$

So,

• $v(2) = 1.9 \, \text{m/s}$
• $v(4) = -17.7 \, \text{m/s}$

Part (b) When does the projectile reach its maximum height?

The projectile reaches its maximum height when the velocity $v(t) = 0$.

Set $v(t) = 0$: $21.5 - 9.8t = 0$ $9.8t = 21.5$ $t = \frac{21.5}{9.8} \approx 2.19 \, \text{seconds}$

So, the projectile reaches its maximum height at approximately $t = 2.19 \, \text{s}$.

Part (c) What is the maximum height?

Substitute $t = 2.19$ back into the height function $h(t)$: $h(2.19) = 2 + 21.5 \times 2.19 - 4.9 \times (2.19)^2$ Calculating this: $h(2.19) = 2 + 47.085 - 23.529$ $h(2.19) \approx 25.56 \, \text{m}$

So, the maximum height is approximately $25.56 \, \text{m}$.

Part (d) When does it hit the ground?

The projectile hits the ground when $h(t) = 0$.

Set $h(t) = 0$: $2 + 21.5t - 4.9t^2 = 0$

This is a quadratic equation in the form $-4.9t^2 + 21.5t + 2 = 0$. Solving this using the quadratic formula: $t = \frac{-21.5 \pm \sqrt{(21.5)^2 - 4 \times (-4.9) \times 2}}{2 \times -4.9}$

Calculating the discriminant: $(21.5)^2 = 462.25$ $4 \times 4.9 \times 2 = 39.2$ $462.25 + 39.2 = 501.45$ $\sqrt{501.45} \approx 22.39$

Now, $t = \frac{-21.5 \pm 22.39}{-9.8}$

For the positive root: $t = \frac{-21.5 + 22.39}{-9.8} \approx \frac{0.89}{-9.8} \approx 4.58 \, \text{s}$

So, it hits the ground at approximately $t = 4.58 \, \text{s}$.

Part (e) With what velocity does it hit the ground?

Substitute $t = 4.58$ into the velocity function $v(t)$: $v(4.58) = 21.5 - 9.8 \times 4.58$ $v(4.58) \approx 21.5 - 44.884 = -23.38 \, \text{m/s}$

So, the velocity when it hits the ground is approximately $-23.38 \, \text{m/s}$ (the negative sign indicates downward direction).

Let me know if you'd like further details on any part, or have any questions!

Here are some related questions you might explore:

1. How would the results change if the initial velocity was different?
2. What effect does the initial height have on the maximum height reached?
3. How does gravity (represented by the $-4.9t^2$ term) influence the projectile's motion?
4. Can we find the time taken for the projectile to fall from its maximum height to the ground?
5. How would air resistance affect the calculations above?

Tip: For any projectile motion problem, always start by identifying initial velocity and the acceleration due to gravity.