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 .

1 answer

  1. 7.45

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