While rising 0.54m at the end of the track, her velocity decreases to V2 that is given by
V1^2 - V2^2 = 2gh = 10.58 m^2/s^2
where V1 = 4.8 m/s
That tells you that
V2 = sqrt(23.04-10.58) = 3.53 m/s
The vertical velocity component leaving the ramp is
V2y = V2*sin48 = 2.62 m/s
How high she rises depends upon the value of V2y.
2gH = (V2y)^2
Solve for H.
a skateboarder moving at v = 4.8 m/s along a horizontal section of a track that is slanted upward by 48° above the horizontal at its end, which is h = 0.54 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.
2 answers
drwls response is the correct way to do this problem.
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