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?

35.2

Homework Answers

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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