When throwing an​ object, the distance achieved depends on its initial​ velocity, v 0​, and the angle above the horizontal at which the object is​ thrown, theta. The​ distance, d, in​ feet, that describes the range covered is given by the equation​ below, where v 0 is measured in feet per second.
d equals StartFraction left parenthesis v 0 right parenthesis squared sine left parenthesis theta right parenthesis cosine left parenthesis theta right parenthesis Over 16 EndFraction
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Part 1
You and your friend are throwing a baseball back and forth. If you throw the ball with an initial velocity of v 0equals80 feet  per​ second, at what angle of​ elevation, to the nearest​ degree, should you direct your throw so that it can be easily caught by your friend located 170 feet​ away?
thetaalmost equals
  
enter your response here degrees 
​(Type your answer in degrees. Use a comma to separate answers as needed. Round to the nearest integer as​ needed.)

Given:
- The distance \( d = 170 \) feet.
- The initial velocity \( v_0 = 80 \) feet per second.
- The formula for the range \( d \) is:

\[
d = \frac{(v_0)^2 \sin(2\theta)}{16}
\]

We need to find the angle \( \theta \) such that the distance \( d = 170 \) feet.

### Step-by-Step Solution

1. **Substitute the known values:**

   \[
   170 = \frac{(80)^2 \sin(2\theta)}{16}
   \]

2. **Simplify the equation:**

   - Calculate \( (80)^2 = 6400 \).
   - Substitute this value:

     \[
     170 = \frac{6400 \sin(2\theta)}{16}
     \]

   - Divide \( 6400 \) by \( 16 \):

     \[
     170 = 400 \sin(2\theta)
     \]

3. **Isolate \( \sin(2\theta) \):**

   \[
   \sin(2\theta) = \frac{170}{400} = 0.425
   \]

4. **Solve for \( 2\theta \):**

   - Take the inverse sine of both sides:

     \[
     2\theta = \arcsin(0.425)
     \]

   - This gives \( 2\theta \approx 25.17^\circ \).

5. **Find \( \theta \):**

   \[
   \theta = \frac{25.17}{2} \approx 12.6^\circ
   \]

### Solution

To the nearest integer, the angle \( \theta \) should be approximately:
\[
\theta \approx 13^\circ
\]

Would you like further details or have any questions?

---

Here are five related questions:
1. How would the angle change if the initial velocity was higher?
2. What would be the effect of increasing the distance \( d \) on the angle \( \theta \)?
3. How does air resistance affect the ideal angle calculated here?
4. What happens to the angle if the target distance is less than 170 feet?
5. How would the formula change if gravity were not constant?

**Tip:** For optimal range in projectile motion without resistance, the ideal angle is usually 45 degrees.

When throwing an object, the distance achieved depends on its initial velocity, v_0, and the angle above the horizontal at which the object is thrown, theta. The distance, d, in feet, that describes the range covered is given by the equation d = (v_0^2 * sin(theta) * cos(theta)) / 16. You and your friend are throwing a baseball back and forth. If you throw the ball with an initial velocity of v_0 = 80 feet per second, at what angle of elevation, to the nearest degree, should you direct your throw so that it can be easily caught by your friend located 170 feet away?

Find the optimal angle to throw a baseball at 80 feet per second to reach a distance of 170 feet. This problem involves key trigonometric concepts and formulas for projectile motion, making it ideal for high school students in Grades 11-12.

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