If the bullet in Problem 65 is traveling at 2,828 feet per second, how long does it take for the bullet to reach a height of 1,000 feet? The bullet was shot out at 45 degtees

To determine the time it takes for the bullet to reach a height of 1,000 feet, given that the bullet was fired at a 45-degree angle with an initial velocity of 2,828 feet per second, we need to break down the problem into components using physics.

### Step 1: Break Down the Velocity into Components
The initial velocity \( v_0 \) can be broken down into horizontal and vertical components:
- Vertical component of velocity: \( v_{0y} = v_0 \sin(\theta) \)
- Horizontal component of velocity: \( v_{0x} = v_0 \cos(\theta) \)

Given:
- \( v_0 = 2,828 \, \text{ft/s} \)
- \( \theta = 45^\circ \)

For \( \theta = 45^\circ \):
- \( \sin(45^\circ) = \cos(45^\circ) = \frac{\sqrt{2}}{2} \)

Thus:
\[ v_{0y} = v_{0x} = 2,828 \times \frac{\sqrt{2}}{2} = 2,828 \times 0.7071 \approx 2,000 \, \text{ft/s} \]

### Step 2: Use the Vertical Motion Equation
The vertical motion of the bullet is governed by the following kinematic equation:
\[ y = v_{0y}t - \frac{1}{2}gt^2 \]

Where:
- \( y = 1,000 \, \text{ft} \) (height)
- \( g = 32.2 \, \text{ft/s}^2 \) (acceleration due to gravity)
- \( t \) is the time in seconds

Substituting the values:
\[ 1,000 = 2,000t - \frac{1}{2}(32.2)t^2 \]

This simplifies to:
\[ 16.1t^2 - 2,000t + 1,000 = 0 \]

### Step 3: Solve the Quadratic Equation
Now, solve the quadratic equation:
\[ 16.1t^2 - 2,000t + 1,000 = 0 \]

We can use the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here:
- \( a = 16.1 \)
- \( b = -2,000 \)
- \( c = 1,000 \)

Substitute the values:
\[ t = \frac{2,000 \pm \sqrt{(-2,000)^2 - 4(16.1)(1,000)}}{2(16.1)} \]

Simplify:
\[ t = \frac{2,000 \pm \sqrt{4,000,000 - 64,400}}{32.2} \]
\[ t = \frac{2,000 \pm \sqrt{3,935,600}}{32.2} \]
\[ t = \frac{2,000 \pm 1,983.81}{32.2} \]

Now calculate the two possible values of \( t \):
1. \( t_1 = \frac{2,000 + 1,983.81}{32.2} \approx \frac{3,983.81}{32.2} \approx 123.7 \, \text{seconds} \)
2. \( t_2 = \frac{2,000 - 1,983.81}{32.2} \approx \frac{16.19}{32.2} \approx 0.5 \, \text{seconds} \)

### Conclusion:
The smaller value of \( t \) is the time it takes for the bullet to reach a height of 1,000 feet on its way up. Thus, the bullet takes approximately **0.5 seconds** to reach a height of 1,000 feet.

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Would you like more details or have any questions? Here are some related questions you might consider:

1. How long does the bullet stay in the air?
2. What is the maximum height reached by the bullet?
3. How far does the bullet travel horizontally before it hits the ground?
4. What would be the time to reach 1,000 feet if the bullet were fired at a different angle?
5. How would air resistance affect the time and height calculations?

**Tip:** When solving projectile motion problems, always start by breaking down the motion into horizontal and vertical components. This simplifies the calculations and helps in understanding the problem better.

Explore how to calculate the time it takes for a bullet traveling at 2,828 feet per second and fired at a 45-degree angle to reach a height of 1,000 feet. This problem involves using projectile motion and quadratic equations in physics.

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