# Skateboarder on a ramp| Physics

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

SOLUTION:

PART A

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