# Freefall with unknown designation of length

1. Aug 31, 2008

### PatrickR.

An object falls a distance h from rest. If it travels 0.42h in the last 1.00 s, find (a) the time and (b) the height of its fall.

so I know a=-g

and I believe I need to use v= v0 + at to some extent.

I also believe that they are trying to say that in the last 1s of the fall it traveled 42% of the total distance.

however because a=-9.8m/s/s and I am given "h" and not m I do not know how to incorporate that into the equation.

Greatly of appreciative of any help.

2. Aug 31, 2008

### loonychune

Don't worry about the signs in the equations you'll have to use since it FALLS h and so you'd be aswell saying everything downward is positive.

The way i've done this is a bit longwinded but... hopefully useful...

You need also the equation,

$$s = ut + \frac{1}{2}at^2,$$
where u is the initial velocity and s the distance fallen.

You have an equation for its overall journey where s = h, u = 0 and a = g.
You have an equation for the last second where s = 0.42h, a = g, t = 1.
You can work out u for the last second of the journey using u = v_0 + at = g(t-1) where t in this equation is the total time and you minus the last second of journey.

The first equation can be used then with the second equation (since we have worked out u from the third equation) to find t (might mean using the quadratic formula), then you use this in the first to find h.

You hadn't a reply yet and i thought it interesting so i had a go myself... just have a go at following what i've said through and hopefully it all works.

3. Sep 1, 2008

### PatrickR.

Loonychune, thank you so much you've been of great help.

so to solve

Eqn1. h=(1/2)at2

Eqn2. .42h=(1/2)(9.8)(1)2 + u

Eqn3. u=(9.8)(t-1)

plugging 3 into 2 we get h= (-4.9 + 9.8t)/.42

plugging 2 into 1 we get 4.9t2 + -23.3333t + 11.6667=0

the only good root is t= 4.19415 s as our total fall time

so plugging t in for eqn1. h=86.21 m