To calculate the height of the bullet, initial velocity, and the angle of firing from the given average distance of $5.7 \, \text{m}$ and time $1.6 \, \text{seconds}$, we need to make some assumptions based on typical projectile motion principles. Let’s assume the motion follows a parabolic trajectory and we can ignore air resistance.

The key formulas we'll use are:

1. Horizontal distance: $d = v_0 \cdot t \cdot \cos(\theta)$ Where:

   • $d = 5.7 \, \text{m}$ (horizontal distance)
   • $v_0$ is the initial velocity
   • $t = 1.6 \, \text{s}$ is the time of flight
   • $\theta$ is the angle of firing

2. Vertical displacement (height): $y = v_0 \cdot t \cdot \sin(\theta) - \frac{1}{2} g t^2$ Where:

   • $g = 9.8 \, \text{m/s}^2$ (gravitational acceleration)
   • $y$ is the vertical height at time $t$

We'll first calculate the initial velocity and the angle, then use these to calculate the height of the bullet. I'll solve these step by step.

Step 1: Calculate Initial Velocity

We can rearrange the formula for horizontal distance to find the initial velocity $v_0$.

$v_0 = \frac{d}{t \cdot \cos(\theta)}$

However, without knowing the angle $\theta$, we will need to find both $v_0$ and $\theta$ simultaneously. Let me calculate that for you.The solution to the system of equations provides two possible results for the initial velocity $v_0$ and angle $\theta$:

1. $v_0 = 8.61 \, \text{m/s}$, $\theta = 1.14 \, \text{radians}$ (approximately $65.5^\circ$)
2. $v_0 = -8.61 \, \text{m/s}$, $\theta = 4.29 \, \text{radians}$

We can discard the negative velocity as it's physically unrealistic in this context, leaving us with:

• Initial velocity: $8.61 \, \text{m/s}$
• Angle of firing: $1.14 \, \text{radians}$ or $65.5^\circ$

Step 2: Calculate the Maximum Height of the Bullet

To find the maximum height, we use the vertical displacement formula. The height occurs when the vertical velocity is zero, which is at half the total time of flight:

$h = v_0^2 \sin^2(\theta) / (2g)$

Let me calculate the maximum height for you.The maximum height of the bullet is approximately $3.14 \, \text{meters}$.

Summary:

• Initial velocity: $8.61 \, \text{m/s}$
• Angle of firing: $65.5^\circ$ (or $1.14 \, \text{radians}$)
• Maximum height of the bullet: $3.14 \, \text{meters}$

Would you like further details or have any questions?

Here are 5 related questions for further exploration:

1. How would the trajectory change if air resistance was included?
2. What is the horizontal range if the angle is increased to $75^\circ$?
3. How would the time of flight change for different initial velocities?
4. How does gravity affect the trajectory of the bullet?
5. What happens to the maximum height if the firing angle is decreased?

Tip: The maximum range of a projectile is achieved at a firing angle of $45^\circ$ in ideal conditions.