How Long Does It Take for an Object to Fall from Height h?

[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.

 

[Kurdt]

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)?

 

[Kurdt]

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?

 

[Kurdt]

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

 

[Kurdt]

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 v_t 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?

 

[Kurdt]

What about other kinematic equations? Do you know any more?

 

[whaler]

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

 

[Kurdt]

There are some basic formula here

https://www.physicsforums.com/showthread.php?t=110015

 

[whaler]

Thank You For All The Help!