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1. The drawing shows a skateboarder moving at 5.4 m/s along a horizontal section of a track that is
1. The drawing shows a skateboarder moving at 5.4 m/s along a horizontal section of a track that isslanted upward by 48° above the horizontal at its end, which is 0.40 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.
Projectilemotion is the curved path motion of a body launched into the air near the Earth's surface and having ahorizontal velocity.Themaximum heightto which the skateboarder rises,H, is approximately0.6 metersabove the end of the track.Reason:Givenparameter are;Initial velocity of theskateboarder, v₁ = 5.4 m/sInclination of the track, above thehorizontal, θ = 48°Heightof the end of the elevated track, h ≈ 0.40 mPath of theskateboarderwhen she leaves the track = Path of a projectileRequired:Maximum heightHto which sherisesabove the end of the track.Solution;From v² = u² - 2·g·h, at the end of the track where;h = 0.40 mu =Initialvelocity = 5.4g = 9.81 m/s²We have;v₂² ≈ 5.4² - 2 × 9.81 × 0.40 = 21.312The velocity at which she leaves the track, v₂ ≈ √(21.312 m²/s²).At themaximum height,H, we have;= 0Therefore, from² =² - 2·g·H, where;² = 2·g·HWhich gives;Therefore;Themaximum heightto which she risesabove the end of the track,H≈ 0.6 m.Learn more aboutprojectile motionhere:brainly.com/question/12125940...