Asked by RoCu

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?

Answers

Answered by Eden
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.).
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