To find the maximum height reached by the projectile, we can use the following kinematic equation:

$v_y^2 = v_{y_0}^2 - 2g(h - h_0)$

where:

• $v_y$ is the final vertical velocity at the maximum height (which is 0, since the projectile stops rising at its highest point),
• $v_{y_0}$ is the initial vertical velocity,
• $g$ is the acceleration due to gravity ($9.81 \, \text{m/s}^2$),
• $h$ is the maximum height,
• $h_0$ is the initial height (which is 0 for a flat firing range).

We can first find the initial vertical velocity component $v_{y_0}$ from the initial speed $v_0 = 37.9 \, \text{m/s}$ and the angle of projection $\theta = 43.4^\circ$:

$v_{y_0} = v_0 \sin(\theta)$

Now, rearranging the kinematic equation to solve for $h$:

$h = \frac{v_{y_0}^2}{2g}$

Step 1: Calculate $v_{y_0}$

$v_{y_0} = 37.9 \, \text{m/s} \times \sin(43.4^\circ)$

Step 2: Substitute $v_{y_0}$ and $g$ into the height equation.

Let me calculate that for you.The maximum height reached by the projectile is approximately $34.6 \, \text{m}$ (to three significant figures).

Would you like more details or have any further questions?

Here are 5 related questions:

1. How do you calculate the total time of flight for this projectile?
2. What is the horizontal distance traveled by the projectile (the range)?
3. How does changing the angle of projection affect the maximum height?
4. How would air resistance alter the calculations for the maximum height?
5. Can you derive the time it takes to reach the maximum height?

Tip: The vertical component of velocity determines how high the projectile will go, while the horizontal component affects how far it will travel.