A small rocket is fired from a launch pad 20 m above the ground with an initial velocity left angle 250 comma 400 comma 500 right angle ​m/s. A cross wind blowing to the north produces an acceleration of the rocket of 2.5 m divided by s squared. Assume the​ x-axis points​ east, the​ y-axis points​ north, the positive​ z-axis is vertical​ (opposite g), and the ground is horizontal.
d. Determine the maximum height of the object.

We are tasked with finding the maximum height of a rocket launched from a point 20 meters above the ground. The given parameters are:

- Initial position of the rocket: \(\mathbf{r}_0 = \langle 0, 0, 20 \rangle\) meters.
- Initial velocity of the rocket: \(\mathbf{v}_0 = \langle 250, 400, 500 \rangle\) meters/second.
- Crosswind acceleration acting on the rocket: \(\mathbf{a}_{\text{wind}} = \langle 0, 2.5, 0 \rangle\) meters/second².
- Gravitational acceleration: \(\mathbf{g} = \langle 0, 0, -9.8 \rangle\) meters/second².

### Step 1: Analyze the motion in the z-direction (vertical)

The motion in the \(z\)-direction is influenced by the initial velocity in that direction and gravity.

The vertical position function for the motion in the \(z\)-direction is given by:
\[
z(t) = z_0 + v_{z0}t - \frac{1}{2} g t^2
\]
Where:
- \(z_0 = 20 \, \text{m}\) (initial height),
- \(v_{z0} = 500 \, \text{m/s}\) (initial vertical velocity),
- \(g = 9.8 \, \text{m/s}^2\) (acceleration due to gravity).

To find the time \(t_{\text{max}}\) when the maximum height is reached, we set the vertical velocity to zero at the peak:
\[
v_z(t) = v_{z0} - g t = 0
\]
\[
t_{\text{max}} = \frac{v_{z0}}{g} = \frac{500}{9.8} \approx 51.02 \, \text{seconds}.
\]

### Step 2: Calculate the maximum height

Substitute \(t_{\text{max}}\) into the equation for \(z(t)\) to find the maximum height:
\[
z_{\text{max}} = z_0 + v_{z0}t_{\text{max}} - \frac{1}{2} g t_{\text{max}}^2
\]
\[
z_{\text{max}} = 20 + 500(51.02) - \frac{1}{2}(9.8)(51.02)^2
\]
Now, let's compute this expression.

The maximum height of the rocket is approximately 12,775.1 meters, and it reaches this height after about 51.02 seconds.

Would you like further details or have any questions?

Here are five related questions to explore:
1. How would the maximum height change if the initial velocity in the z-direction were increased?
2. How long does it take for the rocket to fall back to the ground after reaching maximum height?
3. What is the total time of flight of the rocket?
4. How does the crosswind affect the horizontal displacement of the rocket?
5. What would happen if there were no gravity acting on the rocket?

**Tip:** Always analyze motion in each direction separately when multiple forces or components are involved.

A small rocket is fired from a launch pad 20 m above the ground with an initial velocity ⟨250, 400, 500⟩ m/s. A crosswind blowing to the north produces an acceleration of 2.5 m/s². Assume the x-axis points east, the y-axis points north, the positive z-axis is vertical (opposite g), and the ground is horizontal. Determine the maximum height of the object.

This problem explores the calculation of the maximum height of a rocket launched from 20 meters above the ground with an initial velocity of ⟨250, 400, 500⟩ m/s. Using kinematic equations and considering the effects of gravity, find out how long the rocket takes to reach its peak height of approximately 12,775.1 meters.

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