To find the firing angles that will produce a range of 300 m, we can use the projectile motion formula for range:

$R = \frac{v_0^2 \sin(2\theta)}{g}$

Where:

• $R$ is the range (300 m),
• $v_0$ is the initial speed (70 m/s),
• $\theta$ is the firing angle,
• $g$ is the acceleration due to gravity (approximately 9.8 m/s²).

We can rearrange the formula to solve for $\sin(2\theta)$:

$\sin(2\theta) = \frac{R \cdot g}{v_0^2}$

Now substitute the known values:

$\sin(2\theta) = \frac{300 \cdot 9.8}{70^2}$

Let me calculate that.We have:

$\sin(2\theta) = 0.6$

Now, to find the angles $2\theta$, we take the inverse sine ($\sin^{-1}$) of 0.6:

$2\theta = \sin^{-1}(0.6)$

This will give us two possible values for $2\theta$, since $\sin$ is positive in both the first and second quadrants. Let's calculate both angles for $2\theta$ and then divide them by 2 to get $\theta$.The two possible firing angles that will produce a range of 300 meters are approximately:

• $18.4^\circ$
• $71.6^\circ$

These are the angles that will ensure the projectile reaches the required range.

Would you like further details on the calculations?

Here are 5 follow-up questions to explore more about this:

1. How does changing the initial speed affect the range?
2. What would be the maximum height reached at these angles?
3. How does air resistance impact projectile motion?
4. What is the time of flight for each of these angles?
5. What are the differences in velocity components at these two angles?

Tip: The angle that gives the maximum range in ideal conditions is $45^\circ$, but other angles can yield the same range for different trajectories.