A skateboarder shoots off a ramp with a velocity of 6.8 m/s, directed at an angle of 56° above the horizontal. The end of the ramp is 1.1 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. (a) How high above the ground is the highest point that the skateboarder reaches?
upward component of velocity at takeoff is 6.8 sin 56° = 5.64
h(t) = 1.1 + 5.64t - 4.9t^2
max height reached at t = 5.64/9.8 = 0.5755
h(0.5755) = 2.723
To find the highest point that the skateboarder reaches, we need to analyze the motion of the skateboarder using the given information.
Let's break down the motion into horizontal (x-axis) and vertical (y-axis) components.
First, let's analyze the initial vertical velocity of the skateboarder. The velocity can be broken down into two components: the vertical component and the horizontal component.
Given:
Initial velocity (v) = 6.8 m/s
Launch angle (θ) = 56° above the horizontal
Vertical component (vy):
vy = v * sin(θ)
vy = 6.8 * sin(56°)
To find the highest point, we need to determine when the vertical component of velocity becomes zero. At that point, the skateboarder reaches the maximum height.
Using the kinematic equation for vertical motion:
vy^2 = u^2 + 2as
Since the skateboarder reaches the highest point, the final vertical velocity (vfy) will be zero.
0 = (6.8 * sin(56°))^2 + 2 * (-9.8) * Δy
Where acceleration (a) is -9.8 m/s² (considering the downward direction due to gravity) and Δy is the change in vertical position.
Simplifying the equation:
(6.8 * sin(56°))^2 = 2 * 9.8 * Δy
Now, solve for the change in vertical position (Δy):
Δy = (6.8 * sin(56°))^2 / (2 * 9.8)
Calculating Δy:
Δy = (6.8 * 0.829)^2 / (2 * 9.8)
Δy ≈ 0.993 m
Therefore, the highest point that the skateboarder reaches is approximately 0.993 meters above the ground.