physics question concerning free fall of an object?

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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.

Suggestion by dct6412
[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

What do you think? Answer below!