Posted by Anonymous on .
The drawing shows a skateboarder moving at 6.93 m/s along a horizontal section of a track that is slanted upward by 45.6 ° above the horizontal at its end, which is 0.654 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 = 6.93 m/s[45.6o]
Xo = 6.93*cos45.6 = 4.85 m/s
Yo = 6,93*sin45.6 = 4.95 m/s.
h = (Y^2-Yo^2)/2g = (0-4.95^2)/-19.6 =