A skateboarder moving at 7.18 m/s along a horizontal section of a track that is slanted upward by 48.5 ° above the horizontal at its end, which is 0.741 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 H to which the skateboarder rises above the end of the track, we can use the principles of projectile motion.
The first step is to break down the initial velocity of the skateboarder into its horizontal and vertical components. The horizontal component does not change throughout the motion, while the vertical component changes due to gravity.
Given:
- Initial velocity (v₀) = 7.18 m/s
- Angle of inclination (θ) = 48.5°
- Height of the end of the track (h) = 0.741 m
- Acceleration due to gravity (g) = 9.8 m/s²
The horizontal component of velocity (v₀x) can be found using trigonometry:
v₀x = v₀ * cos(θ)
v₀x = 7.18 m/s * cos(48.5°) ≈ 4.79 m/s
The vertical component of velocity (v₀y) can also be found using trigonometry:
v₀y = v₀ * sin(θ)
v₀y = 7.18 m/s * sin(48.5°) ≈ 5.4 m/s
Now, we can find the time taken for the skateboarder to reach the maximum height. At the maximum height, the vertical component of velocity becomes zero.
Using the equation:
v = v₀y - g * t
0 = 5.4 m/s - 9.8 m/s² * t
Solving for t:
t = 5.4 m/s / 9.8 m/s² ≈ 0.55 s
Now, we can find the maximum height (H) using the equation for vertical displacement:
H = v₀y * t - 0.5 * g * t²
H = (5.4 m/s) * (0.55 s) - 0.5 * (9.8 m/s²) * (0.55 s)²
H ≈ 1.895 m
Therefore, the maximum height to which the skateboarder rises above the end of the track is approximately 1.895 meters.