The drawing shows a skateboarder moving at 6.87 m/s along a horizontal section of a track that is slanted upward by 51.9 ° above the horizontal at its end, which is 0.616 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.
Vo = 6.87m/s.[51.9o].
Yo = 6.87*sin51.9 = 5.41 m/s. = Vertical
component of initial velocity.
Y^2 = Yo^2 + 2g*h = 0.
h = -Yo^2/2g = -6.87^2/-19.6 = 2.41 m.
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.
Step 1: Break down the motion into horizontal and vertical components.
In this case, the skateboarder is moving horizontally along the track, so the initial velocity along the horizontal direction (v_x) is 6.87 m/s, and there is no acceleration in the horizontal direction.
The skateboarder also follows projectile motion after leaving the track, which means there is an initial vertical velocity (v_y) and an acceleration due to gravity (g = 9.8 m/s^2).
Step 2: Determine the initial vertical velocity.
The vertical component of the initial velocity can be calculated using the given information about the angle of the track. We can use trigonometry to find v_y:
v_y = v_initial * sin(theta)
where v_initial is the initial velocity (6.87 m/s) and theta is the angle of the track (51.9°).
v_y = (6.87 m/s) * sin(51.9°)
v_y = 5.248 m/s
Step 3: Calculate the time taken to reach the maximum height.
Using the vertical velocity (v_y) and acceleration due to gravity (g), we can find the time it takes for the skateboarder to reach the maximum height. At the highest point, the vertical velocity will become zero.
v_y = v_initial - g * t
0 = 5.248 m/s - (9.8 m/s^2) * t
Solving for t:
t = 5.248 m/s / (9.8 m/s^2)
t ≈ 0.535 seconds
Step 4: Determine the maximum height (H) using the time calculated.
To find the maximum height (H), we need to know the vertical displacement (Δy) and use the equation:
Δy = v_initial * t + (1/2) * g * t^2
Substituting the values we know:
Δy = 5.248 m/s * 0.535 seconds + (1/2) * (9.8 m/s^2) * (0.535 seconds)^2
Δy ≈ 1.393 meters
So, the maximum height to which the skateboarder rises above the end of the track is approximately 1.393 meters.