During a fireworks display, a shell is shot into the air at an angle of θ degrees above the horizontal. The fuse is timed to 9.3 seconds for the shell just as it reaches its highest point above the ground. If the  horizontal displacement of the shell is 183.2 meters when it explodes, what is the angle θ of the initial velocity ? Use g = 9.8 m/s2.

We can approach this problem by first recognizing that it involves projectile motion, and the shell reaches its maximum height after 9.3 seconds. To find the angle \( \theta \), we need to break down the problem into vertical and horizontal components of the motion.

### Step 1: Vertical motion
The vertical component of the motion helps us understand the relationship between time and the initial vertical velocity. At the highest point, the vertical velocity is zero, and the time to reach this point is 9.3 seconds. Using the equation of motion for vertical velocity:

\[
v_y = v_{y0} - g t
\]

At the highest point, \( v_y = 0 \), so:

\[
0 = v_{y0} - g t
\]

Solving for the initial vertical velocity \( v_{y0} \):

\[
v_{y0} = g t = (9.8 \, \text{m/s}^2)(9.3 \, \text{seconds}) = 91.14 \, \text{m/s}
\]

### Step 2: Horizontal motion
In projectile motion, the horizontal velocity \( v_x \) remains constant because we neglect air resistance. The horizontal displacement \( x \) is given as 183.2 meters, and the total time of flight is twice the time to reach the highest point (since the upward and downward motions are symmetrical). So, the total time of flight is:

\[
T = 2 \times 9.3 \, \text{seconds} = 18.6 \, \text{seconds}
\]

Now, using the equation for horizontal displacement:

\[
x = v_x \times T
\]

Solving for \( v_x \):

\[
v_x = \frac{x}{T} = \frac{183.2 \, \text{m}}{18.6 \, \text{seconds}} \approx 9.85 \, \text{m/s}
\]

### Step 3: Finding the angle \( \theta \)
Now that we know the horizontal and vertical components of the initial velocity, we can find the angle \( \theta \) using trigonometry. The initial velocity components are related to the angle by:

\[
\tan \theta = \frac{v_{y0}}{v_x}
\]

Substituting the known values:

\[
\tan \theta = \frac{91.14 \, \text{m/s}}{9.85 \, \text{m/s}} \approx 9.25
\]

Now, take the inverse tangent to find \( \theta \):

\[
\theta = \tan^{-1}(9.25) \approx 83.8^\circ
\]

### Final Answer:
The angle \( \theta \) of the initial velocity is approximately \( 83.8^\circ \).

---

Would you like more details, or do you have any other questions?

Here are five related questions to deepen your understanding:
1. How would the angle \( \theta \) change if the horizontal displacement were larger?
2. How does the time of flight relate to the initial velocity in projectile motion?
3. What is the significance of the horizontal and vertical components in determining the motion of a projectile?
4. How would air resistance affect the calculation of the angle \( \theta \)?
5. If the explosion occurred before the shell reached its highest point, how would that affect the calculations?

**Tip:** Always keep in mind that in projectile motion, horizontal and vertical motions are independent of each other!

During a fireworks display, a shell is shot into the air at an angle of θ degrees above the horizontal. The fuse is timed to 9.3 seconds for the shell just as it reaches its highest point above the ground. If the horizontal displacement of the shell is 183.2 meters when it explodes, what is the angle θ of the initial velocity? Use g = 9.8 m/s².

Solve a physics problem involving projectile motion: A firework is launched with a timed fuse of 9.3 seconds to its highest point, and the horizontal displacement at explosion is 183.2 meters. Learn how to find the angle of the initial velocity using kinematic equations and trigonometry. Suitable for Grades 11-12.

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