Use the model for projectile motion, assuming there is no air resistance and g = 32 feet per second per second.
Determine the maximum height and range (in ft) of a projectile fired at a height of 7 feet above the ground with an initial speed of 800 feet per second and at an angle of 45° above the horizontal. (Round your answers to three decimal places.) the max height is 5007.000

To solve this projectile motion problem, we can use the standard kinematic equations for motion under constant acceleration, assuming no air resistance and \( g = 32 \, \text{ft/s}^2 \).

### Step 1: Break down the initial velocity
The initial speed \( v_0 = 800 \, \text{ft/s} \) is fired at an angle of \( \theta = 45^\circ \) above the horizontal. We need to break down this velocity into horizontal and vertical components.

- Horizontal velocity: 
  \[
  v_{0x} = v_0 \cos(\theta) = 800 \times \cos(45^\circ) = 800 \times \frac{1}{\sqrt{2}} = 565.685 \, \text{ft/s}
  \]
- Vertical velocity: 
  \[
  v_{0y} = v_0 \sin(\theta) = 800 \times \sin(45^\circ) = 800 \times \frac{1}{\sqrt{2}} = 565.685 \, \text{ft/s}
  \]

### Step 2: Calculate the maximum height
The maximum height occurs when the vertical velocity is zero, i.e., when \( v_y = 0 \). Using the kinematic equation for vertical motion:
\[
v_y^2 = v_{0y}^2 - 2g(h - h_0)
\]
At maximum height, \( v_y = 0 \), so:
\[
0 = v_{0y}^2 - 2g(h_{\text{max}} - h_0)
\]
Solving for \( h_{\text{max}} \):
\[
h_{\text{max}} = h_0 + \frac{v_{0y}^2}{2g} = 7 + \frac{(565.685)^2}{2 \times 32}
\]
\[
h_{\text{max}} = 7 + \frac{319999.936}{64} = 7 + 4999.999 = 5006.999 \, \text{ft}
\]
Rounding to three decimal places, \( h_{\text{max}} = 5007.000 \, \text{ft} \).

### Step 3: Calculate the time of flight
The time of flight can be found by first calculating the time it takes to reach maximum height and then doubling it (since the motion is symmetric).

The time to reach maximum height is given by:
\[
t_{\text{up}} = \frac{v_{0y}}{g} = \frac{565.685}{32} = 17.677 \, \text{seconds}
\]
The total time of flight is then:
\[
t_{\text{total}} = 2t_{\text{up}} = 2 \times 17.677 = 35.354 \, \text{seconds}
\]

### Step 4: Calculate the horizontal range
The range \( R \) is given by the horizontal velocity multiplied by the total time of flight:
\[
R = v_{0x} \times t_{\text{total}} = 565.685 \times 35.354 = 20000 \, \text{ft}
\]

### Final Results:
- **Maximum Height**: \( 5007.000 \, \text{ft} \)
- **Range**: \( 20000.000 \, \text{ft} \)

Do you need further clarification or additional details? Here are five related questions you can explore:

1. How would the maximum height change if the angle of projection is reduced to 30°?
2. What effect would increasing the initial velocity to 1000 ft/s have on the range and maximum height?
3. How would the results change if the projectile is fired from ground level instead of 7 feet above the ground?
4. Can you determine the velocity of the projectile just before it hits the ground?
5. How would adding air resistance alter the calculations of maximum height and range?

**Tip**: The horizontal and vertical components of projectile motion are independent of each other except for the time they take to travel!

Use the model for projectile motion, assuming there is no air resistance and g = 32 feet per second per second. Determine the maximum height and range (in ft) of a projectile fired at a height of 7 feet above the ground with an initial speed of 800 feet per second and at an angle of 45° above the horizontal. (Round your answers to three decimal places.) the max height is 5007.000

Learn how to determine the maximum height (5007.000 ft) and range (20000.000 ft) of a projectile fired at a height of 7 ft above the ground with an initial speed of 800 ft/s at a 45° angle. This problem covers essential concepts in projectile motion, including kinematic equations and trigonometry. Suitable for high school and college physics students.

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