posted by Mary on .
The drawing shows a skateboarder moving at v = 5 m/s along a horizontal section of a track that is slanted upward by 48° above the horizontal at its end, which is h = 0.52 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.
Isn't this best answered by energy relationships?
Launch velocity upward= 5sinTheta
KE of the horizontal motion is constant, so the change in PE is due to the vertical KE.
mgh=1/2 m 25 sin^2Theta.
This is the h above the track the skateboarder rises .