A skateboarder shoots off a ramp with a velocity of 6.4 m/s, directed at an angle of 54° above the horizontal. The end of the ramp is 1.4 m above the ground. Let the x axis be parallel to the ground, the +y direction be vertically upward, and take as the origin the point on the ground directly below the top of the ramp.

Step 1: Resolve the initial velocity into its x and y components.

The initial velocity (vi) of the skateboarder can be resolved into horizontal (vx) and vertical (vy) components using trigonometry.

vx = vi * cosθ
vy = vi * sinθ

where θ is the angle of 54°.

Substituting the given values:
vx = 6.4 m/s * cos(54°)
vx ≈ 6.4 m/s * 0.5878
vx ≈ 3.759 m/s

vy = 6.4 m/s * sin(54°)
vy ≈ 6.4 m/s * 0.8090
vy ≈ 5.1816 m/s

So, the horizontal component of the velocity is approximately 3.759 m/s and the vertical component is approximately 5.1816 m/s.

Step 2: Determine the time taken to reach the ground.

First, we need to find the time taken for the skateboarder to reach the maximum height. At maximum height, vy will be zero.

Using the equation:

0 = vy - gt

where g is the acceleration due to gravity (-9.8 m/s²).

0 = 5.1816 m/s - 9.8 m/s² * t

Solving for t:

t = 5.1816 m/s / 9.8 m/s²
t ≈ 0.5291 s

Since the skateboarder takes the same amount of time to reach the maximum height as it takes to return to the ground, the total time of flight will be twice the time taken to reach the maximum height.

Total time of flight = 2 * 0.5291 s
Total time of flight ≈ 1.0582 s

So, it takes approximately 1.0582 seconds for the skateboarder to reach the ground.

Step 3: Calculate the range (horizontal distance).

The range can be found using the formula:

Range = vx * total time of flight

Substituting the values:

Range = 3.759 m/s * 1.0582 s
Range ≈ 3.9845 m

Therefore, the skateboarder will travel a horizontal distance of approximately 3.9845 meters before landing on the ground.

To analyze the motion of the skateboarder, we can break down the initial velocity into its horizontal and vertical components.

Given:
- Initial velocity magnitude: 6.4 m/s
- Angle above the horizontal: 54°
- Height of the ramp: 1.4 m

First, let's find the horizontal and vertical components of the initial velocity:

Horizontal component:
Vx = V * cos(θ)
Vx = 6.4 m/s * cos(54°)
Vx = 6.4 m/s * 0.5878
Vx ≈ 3.75 m/s

Vertical component:
Vy = V * sin(θ)
Vy = 6.4 m/s * sin(54°)
Vy = 6.4 m/s * 0.8090
Vy ≈ 5.18 m/s

Now, let's analyze the vertical motion of the skateboarder.

Using the second equation of motion for vertical motion:
d = Vyo * t + (1/2) * a * t^2

Since the skateboarder launches at the top of the ramp, the initial vertical velocity (Vyo) is equal to Vy ≈ 5.18 m/s.

The acceleration (a) due to gravity is approximately -9.8 m/s² since it acts downward.

The initial position (d) is 1.4 m above the ground, so the final position will be 0.

0 = 5.18 m/s * t + (1/2) * (-9.8 m/s²) * t^2

Rearranging the equation:

-4.9 t^2 + 5.18 t - 1.4 = 0

To solve this quadratic equation for time (t), we can use the quadratic formula:

t = [-b ± √(b^2 - 4ac)] / 2a

Using a = -4.9, b = 5.18, and c = -1.4, we can substitute these values into the equation:

t = [-5.18 ± √(5.18^2 - 4 * (-4.9) * (-1.4))] / (2 * (-4.9))

Simplifying further:

t = [-5.18 ± √(26.8284 + 27.44)] / (-9.8)
t = [-5.18 ± √(54.2684)] / (-9.8)

Calculating the discriminant (√(54.2684)):

t = [-5.18 ± 7.37] / (-9.8)

This gives us two possible solutions for time.
1. t1 = (-5.18 + 7.37) / (-9.8) ≈ 0.23 seconds
2. t2 = (-5.18 - 7.37) / (-9.8) ≈ -0.24 seconds

Since time cannot be negative in this context, we discard the second solution.
Therefore, the time it takes for the skateboarder to reach the ground is approximately 0.23 seconds.