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A jet pilot diving vertically down at 600 km/hr wants to make a quarter turn with-
out experiencing an acceleration bigger than5g. At what minimum height must the turn
begin? Assume that the speed is constant and that after the quarter turn the plane, moving
horizontally, is at ground level.
The solution just quotes a=v^2/R where R is radius of the turn.
I wondered if could start with r(t)=f(t)i+g(t)j for suitable f,g and then differentiate to find acceleration and so do question?
any clue as to how?Originally Posted by allsmiles
im thinking of maybe if t=a is time to reach ground
r(t)=cos([90-(90t/a))i+sin([90-(90t/a)]j
with |r(t)|= radius of the circle.
v(t)=90/a{sin[90-(90t/a)]i-cos[90-(90t/a)]}
|v(t)|=600
now im getting confused??
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