The range R and the maximum height H of a projectile fired at an inclination thetaθ to the horizontal with initial speed v 0v0 are given by the formulas​ below, where g almost equals 32.2g≈32.2 feet per second per second is the acceleration due to gravity. Complete parts A and B.Upper R equals StartFraction 2 v Subscript 0 Superscript 2 Baseline sine theta cosine theta Over g EndFractionR=
2v20sinθcosθ
g     Upper H equals StartFraction v Subscript 0 Superscript 2 Baseline sine squared theta Over 2 g EndFractionH=
v20sin2θ
2gA. Find the range R if the projectile is fired at an angle of 45 degrees45° to the horizontal with an initial speed of 160160 feet per second.

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.

Learn how to calculate the range (R) of a projectile fired at 45 degrees with an initial speed of 160 feet per second. This problem uses the projectile motion formula and trigonometric identities to find the range. Suitable for high school students studying physics or mathematics.

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