The drawing shows a skateboarder moving at 6.88 m/s along a horizontal section of a track that is slanted upward by 37.3 ° 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 to which the skateboarder rises above the end of the track, we can use the principles of projectile motion. Here's the step-by-step explanation on how to solve this problem:
1. Break down the initial velocity of the skateboarder into its horizontal and vertical components. The horizontal component (Vx) remains constant throughout the motion, while the vertical component (Vy) changes due to gravity.
Vx = V * cos(theta)
Vy = V * sin(theta)
In this case, the initial velocity (V) is given as 6.88 m/s, and the angle (theta) is given as 37.3°.
Vx = 6.88 m/s * cos(37.3°)
= 5.45 m/s
Vy = 6.88 m/s * sin(37.3°)
= 4.09 m/s
2. Determine the time it takes for the skateboarder to reach the highest point of their trajectory. We can use the vertical component of velocity and the acceleration due to gravity.
Use the equation: Vy = Vo + a*t
Where Vy is the final vertical velocity at the highest point (0 m/s), Vo is the initial vertical velocity (4.09 m/s), a is the acceleration due to gravity (-9.8 m/s^2), and t is the time.
0 = 4.09 m/s + (-9.8 m/s^2) * t
Solving for t gives:
t = 0.418 seconds
3. Calculate the maximum height (H) using the vertical component of velocity and the time.
Use the equation: H = Vy * t + (1/2) * a * t^2
Where H is the maximum height, Vy is the initial vertical velocity (4.09 m/s), t is the time (0.418 seconds), and a is the acceleration due to gravity (-9.8 m/s^2).
H = 4.09 m/s * 0.418 s + (1/2) * (-9.8 m/s^2) * (0.418 s)^2
Simplifying the equation gives:
H = 0.876 meters
Therefore, the maximum height to which the skateboarder rises above the end of the track is approximately 0.876 meters.