The drawing shows a skateboarder moving at 5.40 m/s along a horizontal section of a track that is slanted upward by 48.9 ° above the horizontal at its end, which is 0.582 m above the ground. When she leaves the track, she follows the characteristic path of projectile motion. Ignoring friction and air resistance, find the maximum height H to which she rises above the end of the track.

To find the maximum height to which the skateboarder rises above the end of the track, we need to consider the projectile motion of the skateboarder after leaving the track.

First, let's calculate the initial velocity of the skateboarder in the vertical direction when leaving the track. To do this, we split the initial velocity into its horizontal and vertical components. Given that the skateboarder is moving at 5.40 m/s along a horizontal section of the track, the horizontal component of the velocity is 5.40 m/s.

To find the vertical component of the velocity, we can use the trigonometric relationship between the angle and the components of the velocity. The vertical component can be calculated using the formula:

Vertical velocity (V_y) = Velocity (V) x sin(angle)

V_y = 5.40 m/s x sin(48.9 °)

V_y ≈ 3.92 m/s

Now, we can determine the time it takes for the skateboarder to reach the maximum height. We know that the vertical motion is affected by gravity, and the highest point of the skateboarder's trajectory occurs when the vertical velocity becomes zero.

Using the formula for vertical displacement in projectile motion, where the initial velocity is V_y, the acceleration due to gravity is -9.8 m/s^2 (negative for the upward motion), and the maximum height is H:

Vertical displacement (Δy) = (V_y)^2 / (2g)

H = (3.92 m/s)^2 / (2 x 9.8 m/s^2)

H ≈ 0.774 m

Therefore, the maximum height to which the skateboarder rises above the end of the track is approximately 0.774 meters.