a skateboarder moving at v = 5 m/s along a horizontal section of a track that is slanted upward by 48° above the horizontal at its end, which is h = 0.52 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.
This was answered by BobPursley in a separate post.
To find the maximum height to which the skateboarder rises above the end of the track, we can use the principles of projectile motion.
Let's break down the problem and find the solution step by step.
Step 1: Find the initial vertical velocity component (v_y) when the skateboarder leaves the track.
Since the track is slanted at an angle of 48° above the horizontal, we can use trigonometry to find the vertical component of the initial velocity:
v_y = v * sin(48°)
Given that v = 5 m/s, we can calculate v_y:
v_y = 5 * sin(48°) ≈ 3.791 m/s
Step 2: Find the time it takes for the skateboarder to reach the maximum height.
In projectile motion, the time taken to reach the maximum height (t_max) is given by the formula:
t_max = v_y / g
where g is the acceleration due to gravity, approximately 9.8 m/s².
Plugging in the values, we get:
t_max = 3.791 m/s / 9.8 m/s² ≈ 0.387 seconds
Step 3: Find the maximum height (H) using the equation for vertical displacement.
The maximum height reached by the skateboarder can be calculated using the equation:
H = v_y * t_max - 0.5 * g * t_max²
Plugging in the values we already calculated, we get:
H = 3.791 m/s * 0.387 s - 0.5 * 9.8 m/s² * (0.387 s)²
Solving this equation, we find:
H ≈ 0.738 meters
Therefore, the maximum height to which the skateboarder rises above the end of the track is approximately 0.738 meters.