To find the range $R$ of the projectile when it is fired at an angle of $\theta = 45^\circ$ with an initial speed of $v_0 = 160$ feet per second, we use the formula for range:

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

Step 1: Plug in the given values

• $v_0 = 160$ feet per second
• $\theta = 45^\circ$
• $g = 32.2$ feet per second squared

Step 2: Simplify the trigonometric functions

Since $\theta = 45^\circ$, we know: $\sin 45^\circ = \cos 45^\circ = \frac{\sqrt{2}}{2}$

Step 3: Substitute into the formula

$R = \frac{2(160)^2 \sin 45^\circ \cos 45^\circ}{32.2}$

$R = \frac{2(160)^2 \left(\frac{\sqrt{2}}{2}\right) \left(\frac{\sqrt{2}}{2}\right)}{32.2}$

$R = \frac{2(160)^2 \left(\frac{2}{4}\right)}{32.2}$

$R = \frac{160^2}{32.2}$

Step 4: Compute the result

Now, calculate $160^2$ and then divide by $32.2$.

$160^2 = 25600$

$R = \frac{25600}{32.2} \approx 795.03 \text{ feet}$

So, the range $R$ is approximately 795.03 feet.

Would you like the details of the computation, or do you have any other questions?

Here are some related questions to consider:

1. How does the range change if the angle is increased to $60^\circ$?
2. What would the maximum height be for the same projectile at $45^\circ$?
3. How does the initial speed affect the range of the projectile?
4. What happens to the range if the angle is decreased to $30^\circ$?
5. How would you calculate the time of flight for the projectile?

Tip: Remember that the angle $45^\circ$ gives the maximum range for any given initial speed, assuming no air resistance.