# Find the time of its fall

whaler
I have been having some problems with this question:

An object falls from a height h from rest. If it travels a fraction of the total height of 0.5665 in the last 1.00 s, find the time of its fall.

I started by saying h=(0.5)(g)(T^2)
Then I said that 1-.5665h=(0.5)(g)(T-1)^2

I have no idea if this is even remotly right. Please help.

## Answers and Replies

Staff Emeritus
Science Advisor
Gold Member
You have the correct equation. All you need to do is set T=1 and you know that the distance traveled is 0.5665h. Then you can solve for the height put it back in the original equation and solve for total time.

whaler
so that it states

h=(.5)(9.81)(1)?

Staff Emeritus
Science Advisor
Gold Member
so that it states

h=(.5)(9.81)(1)?

you would have $$0.5665 h =\frac{1}{2}g$$

Now rearrange for h.

When you find h put it back into the original equation and solve for t.

whaler
i got an h=8.66
Then i put that in and got a value
t=1.33s

That is not the right answer though...where am i going wrong?

Staff Emeritus
Science Advisor
Gold Member
i got an h=8.66
Then i put that in and got a value
t=1.33s

That is not the right answer though...where am i going wrong?

No worries its my fault. Give me a minute (doing many things at once) and I'll correct my mistake.

whaler
know the feeling...not a problem

Staff Emeritus
Science Advisor
Gold Member
Ok so the equation that describes the motion is:

$$s=h-\frac{1}{2}gt^2$$

Now the question tells you that in the final second the particle falls $s=0.5665h$ therefore,

$$0.5665h = h-v_t-\frac{1}{2}g$$

where vt is the velocity at t-1 seconds. You can work out this velocity and should get a quadratic in h.

whaler
so how would I find the vt though?
would i solve the first equation for h and then plug that into the second?

Staff Emeritus
Science Advisor
Gold Member
What about other kinematic equations? Do you know any more?

whaler
my prof does not teach? have not be given any others yet

whaler
Thank You For All The Help!