To solve the given problems regarding the projectile motion shown in the image, let's analyze the data and apply the equations of motion step by step.

1. Altitude $y_9$ at position 9:

The altitude of the projectile at any time $t$ can be calculated using the vertical motion equation:

$y = v_0 \sin \theta \, t - \frac{1}{2} g t^2$

Where:

• $v_0 = 708 \, \text{m/s}$ (initial velocity),
• $\theta = 36^\circ$ (launch angle),
• $g = 10 \, \text{m/s}^2$ (gravitational acceleration),
• $t$ is the time corresponding to position 9.

From the graph:

• The time interval between each position is $\Delta t = 4.0 \, \text{s}$.
• Therefore, at position 9, $t_9 = 9 \cdot 4 = 36 \, \text{s}$.

Now substitute these values into the vertical motion equation:

$y_9 = (708 \sin 36^\circ)(36) - \frac{1}{2} (10)(36^2)$

We first calculate intermediate terms:

• $\sin 36^\circ \approx 0.5878$,
• $708 \sin 36^\circ = 708 \cdot 0.5878 \approx 415.9 \, \text{m/s}$,
• $\frac{1}{2} g t^2 = \frac{1}{2} (10) (36^2) = 5 \cdot 1296 = 6480 \, \text{m}$.

Now calculate $y_9$:

$y_9 = (415.9)(36) - 6480$ $y_9 = 14972.4 - 6480 \approx 8492.4 \, \text{m}$

Thus:

$y_9 \approx 8492.4 \, \text{m}$

Rounded to the nearest hundredth:

$\boxed{y_9 = 8492.40 \, \text{m}}$

2. Downrange distance $x_7$ at position 7:

The horizontal distance (range) of the projectile at any time $t$ can be calculated using the horizontal motion equation:

$x = v_0 \cos \theta \, t$

Where:

• $v_0 = 708 \, \text{m/s}$,
• $\theta = 36^\circ$,
• $t_7 = 7 \cdot 4 = 28 \, \text{s}$.

First, calculate $\cos 36^\circ$:

• $\cos 36^\circ \approx 0.809$,
• $v_0 \cos 36^\circ = 708 \cdot 0.809 \approx 572.77 \, \text{m/s}$.

Now substitute into the equation for $x_7$:

$x_7 = (572.77)(28)$

Calculate:

$x_7 = 16037.56 \, \text{m}$

Rounded to the nearest hundredth:

$\boxed{x_7 = 16037.56 \, \text{m}}$

Final Answers:

1. Altitude at position 9: $y_9 = 8492.40 \, \text{m}$,
2. Downrange distance at position 7: $x_7 = 16037.56 \, \text{m}$.

Do you need further clarification or details? Let me know!

Here are 5 related questions to expand your understanding:

1. How do you determine the time of flight for the projectile in this case?
2. What is the total horizontal range of the projectile before it hits the ground?
3. How can you find the maximum height reached by the projectile?
4. What is the projectile's velocity at position 9 (both horizontal and vertical components)?
5. How would changing the launch angle affect the range and maximum height of the projectile?

Tip: Remember that projectile motion is split into independent horizontal and vertical components of motion. Use trigonometric functions $\sin$ and $\cos$ to resolve the velocity.