We are given the range formula for a projectile:

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

where:

• $v_0 = 104$ ft/sec (initial velocity)
• $\theta = 26^\circ$ (angle of projection)
• $g = 32.2$ ft/sec² (acceleration due to gravity)

Step 1: Compute $2\theta$

$2\theta = 2 \times 26 = 52^\circ$

Step 2: Compute $\sin 52^\circ$

Using a calculator:

$\sin 52^\circ \approx 0.7880$

Step 3: Compute $v_0^2$

$104^2 = 10816$

Step 4: Compute the Range

$R = \frac{10816 \times 0.7880}{32.2}$

$R = \frac{8523.01}{32.2}$

$R \approx 264.69 \text{ ft}$

Final Answer:

The correct option is C. 264.69 ft.

Would you like a deeper explanation or step-by-step breakdown? Here are five related questions you might find interesting:

1. How does changing the angle $\theta$ affect the range of a projectile?
2. What is the maximum range of a projectile and at what angle does it occur?
3. How would the range change if the initial velocity were doubled?
4. What is the time of flight of the projectile in this problem?
5. How would this problem differ if solved using metric units (m/s² for gravity)?

Tip: The optimal angle for maximum range in projectile motion (without air resistance) is 45°.