Question

A skateboarder in a death-defying stunt decides to launch herself from a ramp on a hill....

A skateboarder in a death-defying stunt decides to launch herself from a ramp on a hill. The skateboarder leaves the ramp at a height of 1.4 m above the slope, traveling 15 m s-1 and at an angle of 30o to the horizontal. The slope is inclined at 45o to the horizontal. With what velocity does the skateboarder land on the slope?

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Answer #1

Dy = Dx tan

and Dy = Doy t + a t^2/2

Dx = 2V^2(tan - tan )(cos)^2 /9.81 = 2x15^2 (tan45 - tan30) (cos30)^2 / 9.81 = 14.54 m/s

t = Dx/ Vos30 = 14.54/ 15cos30 = 1.12 s

so, Dy = Vo Sin30 t + (-a) t^2/2 = 15 x sin45 x 1.12 + (-9.81)x1.12^2/2 = 8.4 - 6.15 = - 2.25 m/s

so, D = sqrt (Dx^2 + Dy^2 ) = sqrt ( 14.54^2 + 2.25^2) = 14.71 m/s

Vx = VoySin30  t + (-a) t = 15 x sin30 x1.12 - 9.81x1.12 = -2.5872 m/s

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