Sunday, February 28, 2010

Fox 253 Solution



Bleaug goes on:

Energy acquired in free fall from rest position is 1/2.mV2=mg(h0-1/2)

Vertical velocity after bounce is V reduced by a factor equal to cos(2a), where a is the angle of tangent to 1/x at B: Vy = V.cos(2a) and tan(a)=-1/4

Highest point reached after bounce is such that: 1/2.mVy2=mg(h-1/2), so h = 1/2 + cos(2a)2.(h0-1/2)

From simple trigonometry we get cos(2a) = 1-tan(a)2/(1+tan(a)2) = 15/17

So h = 1/2 + (15/17)2 . (h0-1/2) = 1/2 + (225/289). 289/2 = 113.

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