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?

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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.