The drawing shows a skateboarder moving at 7.58 m/s along a horizontal section o

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using conservation of energy principle kinetic energy at the bottom of the track = kinetic energy at the height of 0.541 m + potential energy at that height => 1/2 X m X 7.58^2 = 1/2 X m X v^2 + m g X 0.541 cancelling m from both sides we get v = 6.83 m/s so the velocity = 6.83 at an angle of 53.3 deg to the horizontal initial vertical velocity = 6.83 sin 53.3 = 5.476 m/s vertical acceleration = -10 m/s^2 final velocity = 0 m/s v^2 = u^2 + 2as => 0 = 5.476^2 - 2 X 10 X s => s = 1.5 m so the total height above the ground = 1.5 + 0.541 = 2.041 m

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