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.
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.
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