A stone has an initial velocity of 100 fps up to the right at 30 degrees with the horizontal. The components of the acceleration are constant at ax=-4 fps^2 and ay=-20 fps^2. Compute the horizontal distance covered until the stone reaches the point 60 ft below its original elevation?

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To solve this problem, we will use the kinematic equations of motion in two dimensions.

Step 1: Break down the initial velocity into its horizontal and vertical components.
The initial velocity of the stone can be broken down into horizontal and vertical components using trigonometry:
Vx = Vi * cos(theta)
Vy = Vi * sin(theta)

Given:
Vi = 100 fps
theta = 30 degrees

Plugging in these values:
Vx = 100 * cos(30) = 100 * sqrt(3)/2 = 50 * sqrt(3) fps
Vy = 100 * sin(30) = 100 * 1/2 = 50 fps

Step 2: Find the time taken to reach 60 ft below its original elevation.
We know that the vertical displacement is -60 ft (negative because it's below the original elevation) and the acceleration in the vertical direction is -20 fps^2.
Using the equation of motion: Δy = Vyi * t + (1/2) * ay * t^2
Plugging in the values:
-60 = 50 * t + (1/2) * (-20) * t^2
-60 = 50t - 10t^2

Rearranging the equation and setting it equal to zero:
10t^2 - 50t - 60 = 0

Step 3: Solve the quadratic equation.
We can solve this quadratic equation either by factoring or using the quadratic formula. In this case, let's use the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)

Plugging in the values:
t = (-(-50) ± sqrt((-50)^2 - 4 * 10 * (-60))) / (2 * 10)
t = (50 ± sqrt(2500 + 2400)) / 20
t = (50 ± sqrt(4900)) / 20
t = (50 ± 70) / 20

This gives two possible values for t:
t1 = (50 + 70) / 20 = 120 / 20 = 6 seconds (discarding the negative solution since time cannot be negative)

Step 4: Find the horizontal distance covered.
To find the horizontal distance covered, use the equation: Δx = Vx * t
Plugging in the values:
Δx = (50 * sqrt(3)) * 6 = 300 * sqrt(3) ft

Therefore, the horizontal distance covered until the stone reaches the point 60 ft below its original elevation is 300 * sqrt(3) ft.

To find the horizontal distance covered by the stone, we need to determine the time it takes for the stone to reach the point 60 ft below its original elevation and then use that time to calculate the horizontal distance.

Step 1: Resolve the initial velocity into its horizontal and vertical components.
The initial velocity (v₀) can be resolved into its horizontal and vertical components using trigonometry.
Given:
Magnitude of initial velocity (v₀) = 100 fps
Angle with the horizontal (θ) = 30 degrees

The horizontal component (v₀x) can be calculated using:
v₀x = v₀ * cos(θ)

The vertical component (v₀y) can be calculated using:
v₀y = v₀ * sin(θ)

Substituting the given values, we get:
v₀x = 100 fps * cos(30 degrees)
v₀y = 100 fps * sin(30 degrees)

Step 2: Calculate the time it takes for the stone to reach 60 ft below its original elevation.
We can use the equation of motion in the vertical direction to find the time (t) it takes for the stone to reach 60 ft below its original elevation.

The equation of motion in the vertical direction is given by:
y = v₀y * t + (1/2) * ay * t²

Where:
y = displacement in the vertical direction (60 ft = -60 ft)
v₀y = initial vertical velocity (calculated in Step 1)
ay = vertical acceleration (-20 fps²)

Substituting the given values, the equation becomes:
-60 ft = (100 fps * sin(30 degrees)) * t + (1/2) * (-20 fps²) * t²

Rearranging the equation, we get a quadratic equation:
-10 fps² * t² + 50 fps * t + 60 ft = 0

Step 3: Solve the quadratic equation to find the time.
Solving the quadratic equation can be done by factoring or using the quadratic formula.

In this case, let's use the quadratic formula:
t = (-b ± sqrt(b² - 4ac)) / 2a

Where:
a = -10 fps²
b = 50 fps
c = 60 ft

Substituting the values, we get:
t = (-50 fps ± sqrt((50 fps)² - 4 * (-10 fps²) * 60 ft)) / (2 * -10 fps²)

Simplifying further, we get:
t = (-50 fps ± sqrt(2500 fps² - (-2400 fps²))) / (-20 fps²)
t = (-50 fps ± sqrt(4900 fps²)) / (-20 fps²)
t = (-50 fps ± 70 fps) / (-20 fps²)

Using the positive root (since time cannot be negative), we get:
t = (-50 fps + 70 fps) / (-20 fps²)
t = 1 s

Therefore, the stone takes 1 second to reach 60 ft below its original elevation.

Step 4: Calculate the horizontal distance covered using the time.
Now that we have the time (t) it takes for the stone to reach 60 ft below its original elevation, we can calculate the horizontal distance (d) covered using the horizontal component of velocity (v₀x) and the time (t) given by:
d = v₀x * t

Substituting the values, we get:
d = (100 fps * cos(30 degrees)) * 1 s

Calculating further, we get:
d ≈ 86.60254038 ft

Therefore, the horizontal distance covered by the stone until it reaches the point 60 ft below its original elevation is approximately 86.6 ft.