Question

A skateboarder starts at point A in the figure(Figure 1) and rises to a height of 2.64 m above the top of the ramp at point B.What was the skateboarder's initial speed at point A?

Answer 1

assuming mechanical energy is conserved

K.E=P.E
\(\dfrac{1}{2}mv^2=mgh\)

\(\dfrac{1}{2}v^2=(9.8~m/s^2)(2.64~m)\)

\(v=\sqrt{2*(9.8~m/s^2)(2.64~m)}=7.2~m/s\)