Question

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 <b>above</b> the track the skateboarder rises .

Answer 1

7.45