Question

A skateboarder starts up a 1.0-m-high, 30∘ ramp at a speed of 5.2 m/s . The skateboard wheels roll without friction. At the top, she leaves the ramp and sails through the air.

Part A

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

Answer 1

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

conservation of energy gives us
initial KE = final KE + final PE
which after dividing through by mass/2 gives
Vi² = Vf² + 2gh
(5.2 m/s)² = Vf² + 2*9.8m/s²*1m
Vf = 2.72 m/s

Now we have the launch velocity, and we know the launch angle = 30º
Vertically, Vy = Vf*sinΘ = 2.72 m/s * sin30º = 1.36 m/s
s = So + Vy*t + ½at²
0 = 1m + 1.36 m/s*t - 4.9m/s²*t²

4.9*t² - 1.36*t - 1 = 0
This quadratic has roots at t = -0.33 s ← not possible
and t = 0.61 s ← time of flight

Then horizontally we have
x = Vx*t = Vf*cosΘ*t = 2.72 m/s * cos30º * 0.61 s = 1.44 m