Physics: A skateboarder starts up a 1.0-m-high, 30∘ ramp at a speed of 7.5 m/s .?

A skateboarder starts up a 1.0m high, 30∘ ramp at a speed of 7.5 m/s . The skateboard wheels roll without friction. At the top, she leaves the ramp and sails through the air.

How far from the end of the ramp does the skateboarder touch down?

1 Answer

  • Strategy: use conservation of energy to find the launch velocity, then basic kinematics.

    CoE gives us

    initial KE = final KE + final PE

    which after dividing through by mass/2 gives

    Vi² = Vf² + 2gh

    (7.5m/s)² = Vf² + 2*9.8m/s²*1m

    Vf = 6.05 m/s

    Now we have the launch velocity, and we know the launch angle = 30º

    Vertically, Vy = Vf*sinΘ = 6.05m/s * sin30º = 3.03 m/s

    s = So + Vy*t + ½at²

    0 = 1m + 3.03m/s*t - 4.9m/s²*t²

    This quadratic has roots at t = -0.24 s ← not possible

    and t = 0.86 s ← time of flight

    Then horizontally we have

    x = Vx*t = Vf*cosΘ*t = 6.05m/s * cos30º * 0.86s = 4.5 m ◄

    Hope this helps!

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