# One Dimensional Motion help

1. Aug 24, 2009

### pointintime

1. The problem statement, all variables and given/known data

A toy rocket passes by a 2.0 m-high window whose sill is 10.0 m above the ground. The rocket takes .15 s to travel the 2.0 m height of the window. What was the launch speed of the rocket, and how high will it go? Assume the propeltant is burned very quickly at blastoff.

2. Relevant equations

def of average velocity = t^-1 (X - Xo)
def of acceleration = t^-1 (V -Vo)
def of average velocity (velocity remains constant) = 2^-1 (V + Vo)

V = Vo + at

X = Xo + vo t + 2^-1 a t^2

V^2 = Vo^2 + 2a (X- Xo)

3. The attempt at a solution

I calculated the velocity needed to accelerate 2.0 m in that time to be 14.07 s^-1 m
using this information, reaching that velocity in 10 m it would have to have an intial velocity from the ground of 19.8 s^-1 m

Sense this is problem 65 the answer is in the back of the book

19.8 s^-1 m; 20.0 m

I do not know how to solve for the how height it will go because all we know is the Vo and the acceleration...

we also know that the velocity at the max height is zero
so I rearanged this equation

def of a = t^-1 (V - Vo)
for t
t = (Vo + a)^-1
sense there is no V and got .03378 s not only is that wrong but the units didn't come out to seconds so I don't know how to find the height...

Also another question when deriving this equation

X = Xo + Vo t + 2^-1 a t^2

we do the following

def of average velocity = t^-1 (X - Xo)

solve for X

X = Vt + Xo
were V in that eqation is average...

then

def of average velocity (v is constant no acceleration) = 2^-1 (V + Vo)

plug in

X = Xo + 2^-1 (V + Vo) t

def of a = t^-1 (V - Vo)
solve for V

V = Vo + at

plug and chug

X = Xo + 2^-1 (at + Vo + Vo)t
X = Xo + 2^-1 (2Vo + at)t
X = Xo + Vo t + 2^-1 a t^2

so my question is we derrived that equation using this formula
def of average velocity (v is constant no acceleration) = 2^-1 (V + Vo)
which is just the def of aritmetic average between two points sense it's linear and the velocity remaind constant with no acceleration but what I don't get is we derrived this equation using a equation that assumed there was no acceleration so how come are derrived equation has an acceleration in it??? I thought we assuemd there was none... This has always struck me as odd...

ThANKS!!!

2. Aug 24, 2009

### kuruman

In this problem, the acceleration is known. What could it be since you are told that all the fuel is gone by the time the rocket reaches the window? Start from there.

3. Aug 24, 2009

### pointintime

wait so all the fule is empty at 12 m in the air???

4. Aug 24, 2009

### kuruman

Yes, all the fuel is gone very quickly, assume instantly. What is the acceleration of the rocket then?

5. Aug 24, 2009

### pointintime

i still don't know how to go about doing it because there would still be a Vo and a acceleration at the top of the window

6. Aug 24, 2009

### kuruman

Sure there will be. This problem is like throwing a rock straight up in the air. What is the acceleration of the rock as it is moving up? Neglect air resistance.