Question 1175536: A ball is thrown a projectile motion. It is known that the horizontal distance (range) the ball can travel is given by R = v^2/g sin 2x, where r is the range (in feet), v^0 is the initial speed (in/ft/s), x is the angle of elevation the ball is thrown, and g=32ft/s^2 is the acceleration due to gravity.
a. Express the new range in terms of the original range when an angle x (0
b. If a ball travels a horizontal distance of 20 ft when kicked at an angle of x with initial speed 20 √2ft/s, find the horizontal distance it can travel when you double x.
Hint: use result of item (a)
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! 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.**
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