Physics question concerning free fall of an object?

so i am really stumped on this question because im not sure how to write an equation for this since the distance is a function of time and the distance is given as a fraction in comparison of time...





An object falls from a height h from rest. If it travels a fraction of the total height of 0.6617 in the last 1.00 s, find the time of its fall.Physics question concerning free fall of an object?
[EDIT - The two guys below made a classic mistake. They have assumed the initial velocity after 1 second is zero. It is not. The object started from zero velocity at some time earlier at a higher starting position. As such, their answers will be wrong]





The general equation for motion subject to a uniform acceleration due to gravity is:


y = -1/2*g*t^2 + vo*t + yo.


where y is the variable of height, yo is the initial position, vo is the initial velocity (positive velocities are upwards), and g is 9.8m/s^2. The first term is negative because gravity is acting downwards.





vo = 0 since the object falls from rest.


yo = h since that's the total height it fell.





y = -1/2*g*t^2 + h





When y = 0, t = T where T is the total time of the fall.


0 = -1/2*g*T^2 + h


h and T are both variables that are unknown.





You know that in the last second it fell a distance of 0.6617*h. That means after T-1 seconds it was at a height of 0.6617*h.





0.6617*h = -1/2*g*(T-1)^2 + h





You now have two equations


h = 1/2*g*T^2


and


0.3383*h = 1/2*g*(T-1)^2





Solve those two equations


0.3383*(1/2*g*T^2) = 1/2*g*(T-1)^2


0.3383*T^2 = (T-1)^2


0.3383 = ((T-1)/T)^2


(T-1)/T = 0.58


T-1 = 0.58T


1 = 0.42T


T = 2.39





So T = 2.38 seconds and h = (1/2*g*T^2) = 28.0 meters





Let's check these numbers against the general equation


y = -1/2*g*t^2 + h





What's the position after T-1 seconds


y = -1/2*g*(T-1)^2 + 27.78


y = 18.53





After 1.38 seconds, the object is at a height of 18.53 meters which is 0.6618*h.





I do believe T = 2.39 seconds