Assuming that the ramp is frictionless, at the top of the ramp:
KE = PE ==> (1/2)mv^2 = mgh.
Solving this v gives:
(1/2)v^2 = gh, as the mass cancels out
==> v = √(2gh), by multiplying both sides by 2 and square-rooting both sides
= √[2(9.8)(2.64)]
= 7.19 m/s (to 3 s.f.).
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?
1 answer
1 answer