Free fall motion problem ?!?

The problem I was given was this:

A person observes that a falling object takes 1.5 seconds to reach the
ground from a point measured at a distance that is 1/3 of it's inital
height. Given that acceleration was a constant 9.8 m/s^2, the inital
velocity was 0 m/s, what was the initial height?

Thanks for any help!

"Mike" <[Only registered users see links. ]> wrote in message
news:yQRRb.3720$u_6.1394@lakeread04...

s = (1/2 a*t^2

If this isn't enough to help you, read the chapter.

Do your own homework.


Tom Davidson
Richmond, VA

Thanks for your try, but its not that simple. The problem is slightly more
complicated then using the position formula. Notice that the time given
starts at the point two thirds of the total distance the object falls so it
already has a velocity when the time starts which is unknown. The total
distance is also an unknown.

What I am having difficulty understanding is the relationship between the
formulas needed to get the answer, not solving simple single variable
equations.

"tadchem" <[Only registered users see links. ]> wrote in message
news:[Only registered users see links. ]...

>

May I suggest you draw a graph of the motion, time horizontal, position
vertical? You already know the shape of the graph, and now you know something
about a distance that's h/3. You know it's 1.5 seconds from h =0. Isn't it
just a question of looking at the picture, and knowing a couple of points on
the curve, doing a bit of substitution and solving two equations for h?

Strain your brain is the right answer, but as mine needs rust shaken off it,
and with no guarantees that I have succeeded:

T = total time falling
2/3 H is where measurement starts

H = a*T*T/2
2*H/3 = a*(T-1.5)*(T-1.5)/2

eliminate T

David F. Cox

"Mike" <rownder@hotmail.com> wrote in message
newskZRb.5037$u_6.3796@lakeread04...
more
it

>The problem I was given was this:


It's approximately 66.15 meters. If you want to know why send me an e-mail at
[Only registered users see links. ]

Jason

"Jason" <[Only registered users see links. ]> wrote in message
news:[Only registered users see links. ].com...
e-mail at

I would say that that's *very* approximate indeed. Better
recheck your work.

"Mike" <rownder@hotmail.com> wrote in message
newskZRb.5037$u_6.3796@lakeread04...

The problem is simple, but not *TOO* simple. Its purpose is to try to get
you to think. The formulae you are learning are tools. You must learn how
to use them.

it

Not all gravity problems start at x=0 and t=0. Take the advice of David F.
Cox and *draw yourself a diagram*.


Think of the s as _distance traveled from a standing start_ and the t as
_elapsed time_...

Once again (with typo corrected),

s = (1/2)*a*t^2

Get that brain of yours out of the box, dust it off, and fire it up.


Tom Davidson
Richmond, VA

"tadchem" <[Only registered users see links. ]> wrote in message
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snip

F.

good advice, but not mine

David F. Cox

"David F. Cox" <[Only registered users see links. ]> wrote in message
news:17oSb.26$[Only registered users see links. ].net...

<snip>


It was the advice of tony. My apologies to both of you.


Tom Davidson
Richmond, VA