<p>The drawing shows a skateboarder moving at v = 4.6 m/s along a horizontal sec

Question

<p>The drawing shows a skateboarder moving at v = 4.6 m/s along a horizontal section of a track that is slanted upward by 48&#176; 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.</p>
<p>I can't insert the diagram, but can someone give me the correct answer please<br /><br /></p>

Explanation / Answer

The speed of the skater v = 4.6 m/s The height reached by the skater is h = 0.52 m Let u be the velocity of the skater at the end of the ramp Then by the conservation of the mechanical energy 0.5mv2 = 0.5mu2 + mgh 0.5*(4.6 m/s)2 = 0.5*u2 + (9.8 m/s2)(0.52 m) u = 3.31 m/s Now the skater represemts a projectile orojected eith speed u at angle with the horizontal The maximum height reached by skater is H = u2sin2/2g H = (3.31 m/s)2sin248/2(9.8 m/s2) H = 0.309 m above the end of the track

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