Skateboarder on a ramp| Physics

A skateboarder starts up a 1.0-m-high, 30° ramp at a speed of 6.9 m/s. The skateboard wheels roll without friction. At the top, she leaves the ramp and sails through the air.

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

SOLUTION:

PART A

ANSWER: 3.7 m

This is a two-part problem. Use an s-axis parallel to the slope for the first part, regular xy-coordinates for the second. The skateboarder’s final velocity at the top of the ramp is her initial velocity as she becomes airborne. 

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Without friction, the skateboarder’s acceleration is a_o=-gsin\:30^{\circ} =-4.90\:m/s^2. The length of the ramp is s_1=\left(1.0\:m\right)/sin\:30^{\circ} =2.0\:m. We can use kinematics to find her speed at the top of the ramp:

\left(v_1\right)^2=\left(v_o\right)^2+2a_o\left(s_1-s_o\right)

v_1=\sqrt{\left(6.9\right)^2+2\left(-4.9\right)\left(2.0\right)}=5.29\:m/s

This is the skateboarder’s initial speed into the air, giving her velocity components v_{1x}=v_1cos\:30^{\circ} =4.58 and v_{1y}=v_1\:sin\:30^{\circ} =2.65\:m/s.  We can use the y-equation of projectile motion to find her time in the air:

y_2=y_1+v_{1y}t_2+\frac{1}{2}a_{1y}\left(t_2\right)^2

\:0=1+2.65t_2-4.9\left(t_2\right)^2

This quadratic equation has roots -0.256 s and 0.797 s. The x-equation of motion is thus

x_2=x_1+v_{1x}t_2=0+\left(4.58\right)\left(0.797\right)=3.7\:m

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