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# Free fall motion problem ?!?

## Free fall motion problem ?!? - Physics Forum

### Free fall motion problem ?!? - Physics Forum. Discuss and ask physics questions, kinematics and other physics problems.

#1
01-28-2004, 03:55 PM
 Mike Guest Posts: n/a
Free fall motion problem ?!?

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!

#2
01-28-2004, 11:50 PM
Free fall motion problem ?!?

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

s = (1/2 a*t^2

Tom Davidson
Richmond, VA

#3
01-29-2004, 12:27 AM
 Mike Guest Posts: n/a
Free fall motion problem ?!?

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.

news:[Only registered users see links. ]...

#4
01-29-2004, 07:49 AM
 tony Guest Posts: n/a
Free fall motion problem ?!?

>

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?
#5
01-29-2004, 12:33 PM
 David F. Cox Guest Posts: n/a
Free fall motion problem ?!?

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
more
it

#6
01-29-2004, 05:23 PM
 Jason Guest Posts: n/a
Free fall motion problem ?!?

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

Jason
#7
01-29-2004, 05:47 PM
 Greg Neill Guest Posts: n/a
Free fall motion problem ?!?

"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

#8
01-29-2004, 11:56 PM
Free fall motion problem ?!?

"Mike" <rownder@hotmail.com> wrote in message

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

#9
01-30-2004, 06:46 AM
 David F. Cox Guest Posts: n/a
Free fall motion problem ?!?

news:[Only registered users see links. ]...

snip

F.

David F. Cox

#10
01-31-2004, 02:28 AM
Free fall motion problem ?!?

"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

 Tags fall , free , motion , problem

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