A skateboarder shoots off a ramp with a velocity of 6.4 m/s, directed at an angle of 61° 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.
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To solve this problem, we need to break down the initial velocity into its x and y components.
The x component, Vx, can be found using the formula Vx = V * cos(θ), where V is the magnitude of the initial velocity and θ is the angle above the horizontal.
Vx = 6.4 m/s * cos(61°)
Vx = 6.4 m/s * 0.4848
Vx ≈ 3.10 m/s
The y component, Vy, can be found using the formula Vy = V * sin(θ).
Vy = 6.4 m/s * sin(61°)
Vy = 6.4 m/s * 0.8746
Vy ≈ 5.60 m/s
Now we can analyze the motion of the skateboarder in the x and y directions separately.
In the x direction, the skateboarder moves with a constant velocity, so the equation for the horizontal distance traveled can be calculated using the formula:
x = Vx * t
We need to find the time it takes for the skateboarder to reach the ground. Since the skateboarder starts at a height of 1.4 m above the ground, we can use the equation for a free-falling object in the y direction:
y = (1/2) * g * t^2 + Vy * t + Δy
where g is the acceleration due to gravity (approximately 9.8 m/s^2), t is the time taken, and Δy is the initial displacement (1.4 m).
Setting y equal to zero (since the skateboarder reaches the ground), we can solve for t:
(1/2) * g * t^2 + Vy * t + Δy = 0
Simplifying the equation, we get:
(1/2) * 9.8 m/s^2 * t^2 + 5.60 m/s * t + 1.4 m = 0
Now we can solve this quadratic equation using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
where a = (1/2) * g, b = Vy, and c = Δy.
Plugging in the values:
t = (-5.60 m/s ± √((5.60 m/s)^2 - 4 * (0.5 * 9.8 m/s^2 * 1.4 m))) / (2 * 0.5 * 9.8 m/s^2)
Simplifying further, we get:
t = (-5.60 m/s ± √(31.36 m^2/s^2 - 27.44 m^2/s^2)) / (9.8 m/s^2)
t = (-5.60 m/s ± √(3.92 m^2/s^2)) / (9.8 m/s^2)
t = (-5.60 m/s ± 1.98 m/s) / (9.8 m/s^2)
Now we have two possible values for t, as indicated by the ± symbol. We can calculate both values:
t1 = (-5.60 m/s + 1.98 m/s) / (9.8 m/s^2)
t1 ≈ 0.40 s
t2 = (-5.60 m/s - 1.98 m/s) / (9.8 m/s^2)
t2 ≈ -0.41 s
Since time cannot be negative, we discard the second value.
Therefore, it takes approximately 0.40 seconds for the skateboarder to reach the ground.