Vo = 5.71m/s[49.8o]
Yo = 5.71*sin49.8 = 4.36 m/s.
Y^2 = Yo^2 + 2g*h = 0 @ max. ht.
h = (Y^2-(Yo^2))/2g
h = (0-(4.36^2))/-19.6 = 0.970 m.
The drawing shows a skateboarder moving at 5.71 m/s along a horizontal section of a track that is slanted upward by 49.8 ° above the horizontal at its end, which is 0.516 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.
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