The drawing shows a skateboarder moving at 6.08 m/s along a horizontal section o

Question

The drawing shows a skateboarder moving at 6.08 m/s along a horizontal section of a track that is slanted upward by 46.9 ° above the horizontal at its end, which is 0.725 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 whioch she rises above the end of the track.

Explanation / Answer

First we find speed of take off using energy method: .5 * mass * 6.08^2 - mass* 9.81 * 0.725 = .5 * mass * u^2 => u = 22.74 m/s h = u sin (angle) / 9.81 = 22.74 * sin 46.9 ° / 9.81 = 1.692 m Total height = .725 + 1.692 = 2.417m

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