A projectile is fired with an initial speed of 80m/s at an angle of 35 degrees above the horizontal on a long flat firing range Determine the maximum height reached by the projectile total time and how far away the objects hits the ground

To solve the problem of a projectile fired at an angle, we need to break down the motion into horizontal and vertical components and apply kinematic equations.

### Given Data:
- Initial speed (\(v_0\)) = 80 m/s
- Angle of projection (\(\theta\)) = 35 degrees
- Gravitational acceleration (\(g\)) = 9.8 m/s²

### 1. Break the initial velocity into components:
- Horizontal component (\(v_{0x}\)) = \(v_0 \cdot \cos(\theta)\)
- Vertical component (\(v_{0y}\)) = \(v_0 \cdot \sin(\theta)\)

### 2. Maximum Height (\(H_{max}\)):
At the maximum height, the vertical velocity is zero. Using the kinematic equation:

\[
v_y^2 = v_{0y}^2 - 2gH
\]
Since \(v_y = 0\) at maximum height, we get:
\[
H_{max} = \frac{v_{0y}^2}{2g}
\]

### 3. Total Time of Flight (\(t_{total}\)):
The total time of flight can be found by considering the motion in the vertical direction. The time taken to reach the maximum height is:

\[
t_{up} = \frac{v_{0y}}{g}
\]

Since the time to go up is equal to the time to come down, the total time is:

\[
t_{total} = 2 \cdot t_{up} = \frac{2v_{0y}}{g}
\]

### 4. Range of the Projectile (\(R\)):
The horizontal range is the total distance the projectile travels before hitting the ground. This can be calculated as:

\[
R = v_{0x} \cdot t_{total}
\]

Now, let's calculate the values.

#### Step-by-Step Calculation:

First, calculate the horizontal and vertical components of the velocity:
- \(v_{0x} = 80 \cdot \cos(35^\circ)\)
- \(v_{0y} = 80 \cdot \sin(35^\circ)\)

Let's compute these values.### Results:

1. **Horizontal velocity** (\(v_{0x}\)) = 65.53 m/s
2. **Vertical velocity** (\(v_{0y}\)) = 45.89 m/s
3. **Maximum height** (\(H_{max}\)) = 107.43 m
4. **Total time of flight** (\(t_{total}\)) = 9.36 seconds
5. **Range** (\(R\)) = 613.68 m

The projectile reaches a maximum height of 107.43 meters, spends 9.36 seconds in the air, and hits the ground 613.68 meters away from the launch point.

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Do you want more details on any of these calculations or have any questions?

### Related Questions:
1. How would the maximum height change if the initial velocity were increased?
2. How would the total time of flight change if the firing angle were 45 degrees?
3. What would happen if the projectile were fired from a height above ground level?
4. How does air resistance affect the range of the projectile?
5. How could you calculate the speed of the projectile at any point during its flight?

**Tip:** Always break down projectile motion into horizontal and vertical components to simplify the analysis.

A projectile is fired with an initial speed of 80 m/s at an angle of 35 degrees above the horizontal on a long flat firing range. Determine the maximum height reached by the projectile, total time, and how far away the object hits the ground.

Solve a projectile motion problem where a projectile is fired with an initial speed of 80 m/s at a 35-degree angle. Calculate the maximum height (107.43 meters), total flight time (9.36 seconds), and range (613.68 meters) using kinematic equations and trigonometry. Suitable for Grades 10-12.

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