Thread: can i do this using postion vectors?

A jet pilot diving vertically down at 600 km/hr wants to make a quarter turn with-
out experiencing an acceleration bigger than

5g. 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?

looks good to me!

Originally Posted by allsmiles

looks good to me!

any clue as to how?

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??