Question 1175536
Let's tackle this projectile motion problem.

**a) Expressing the New Range in Terms of the Original Range**

* **Original Range:** R = (v^2 / g) * sin(2x)
* **New Angle:** 90° - x

Let's find the new range, R', with the new angle:

* **New Range:** R' = (v^2 / g) * sin(2(90° - x))
* R' = (v^2 / g) * sin(180° - 2x)

Now, we use the trigonometric identity: sin(180° - θ) = sin(θ)

* R' = (v^2 / g) * sin(2x)

Comparing this to the original range, R:

* R' = R

**Therefore, the new range is the same as the original range when the angle is changed from x to 90° - x.**

**b) Finding the Horizontal Distance When Doubling x**

1.  **Find the original angle x:**
    * We're given: R = 20 ft, v = 20√2 ft/s, g = 32 ft/s²
    * R = (v² / g) * sin(2x)
    * 20 = ((20√2)² / 32) * sin(2x)
    * 20 = (800 / 32) * sin(2x)
    * 20 = 25 * sin(2x)
    * sin(2x) = 20 / 25 = 4 / 5 = 0.8
    * 2x = arcsin(0.8)
    * 2x ≈ 53.13°
    * x ≈ 26.565°

2.  **Find the new angle 2x:**
    * The new angle is 2x, which we already found to be approximately 53.13°.

3.  **Find the new range R'':**
    * R'' = (v² / g) * sin(2(2x))
    * R'' = (v² / g) * sin(4x)
    * We know 2x ≈ 53.13°, so 4x ≈ 106.26°
    * R'' = ((20√2)² / 32) * sin(106.26°)
    * R'' = 25 * sin(106.26°)
    * R'' ≈ 25 * 0.96
    * R'' ≈ 24 ft

**Therefore, the horizontal distance the ball can travel when you double x is approximately 24 feet.**