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 Homework Statement
 An elevator car whose floortoceiling distance is equal to ##2.7m## starts ascending with constant acceleration ##1.2 m/s^2##. ##2.0 s## after the start a bolt begins falling from the ceiling of the car. Find the bolt's free fall time.
 Homework Equations

##y(t)=\frac{1}{2}at^2+v_0t+y_0##
##l=2.7m## and ##w=1.2m/s^2##
This is my wrong attempt, the bolt's equation for ##t\geq 2## would be ##y_b(t)=\frac{1}{2}g(t2)^2+e(2)+l## where ##e(t)=\frac{1}{2}wt^2## the position of the elevator's floor in the absolute frame of reference.
##d(t)=\frac{1}{2}(w+g)t^2+2gt+2(wg)+l## the distance between the bolt and the elevator.
I get ##\Delta=4g^2+(2g+2w)(2w2g+l)## and so ##t=\frac{2g\pm\sqrt{4g^2+(2g+2w)(2w2g+l)}}{(g+w)}## which yields a wrong answer.
My mistake is probably in the bolt's position equation, I don't see how it's wrong though.
The correct answer is ##0.7s##.
Thank you!
##d(t)=\frac{1}{2}(w+g)t^2+2gt+2(wg)+l## the distance between the bolt and the elevator.
I get ##\Delta=4g^2+(2g+2w)(2w2g+l)## and so ##t=\frac{2g\pm\sqrt{4g^2+(2g+2w)(2w2g+l)}}{(g+w)}## which yields a wrong answer.
My mistake is probably in the bolt's position equation, I don't see how it's wrong though.
The correct answer is ##0.7s##.
Thank you!
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