Question

As shown in the figure below, a skateboarder start

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Answer #1

His initial speed at point A which will be given as :

using conservation of energy, we have

\DeltaK.E = \DeltaP.E

(1/2) m v02 = m g h

v0 = \sqrt{}2 g h

where, g = acceleration due to gravity = 9.8 m/s2

h = maximum height above the top of ramp = 2.13 m

then, we get

v0 = \sqrt{}2 (9.8 m/s2) (2.13 m)

v0 = \sqrt{}41.748 m2/s2

v0 = 6.46 m/s

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