# Can some one just check my work

1. Oct 29, 2009

### robair

can some one just check my work plz

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

model rocket of mass .250 kg is launched vertically with an engine that is ignited at time= 0, the engine provides 20 N.s impulse by firing for 2 sec . upon reaching its maximum hight the rocket deploys a parachute and then desends vertically to the ground.

a) find acceleration during the 2sec firing
b) what will be the max hight
c) at what time after t=0 will the max hight be reached

2. Relevant equations

impulse or change in momentum = F x t ..... sumFy = ma = F*engine* - F *gravity* or mg
change in y = v inital x t + 1/2 a t^2

3. The attempt at a solution

using first equation---> F t = change p ---> F 2 = 20 --> F= 10 N
then second equation F *motor* = ma + mg ---> after you solve for F*motor that is *
10=(.250)a+(.25)(9.81)----> a= 30.19 m/s^2

that was for part A of the question and for part C

i found velocity using p=mv final - mv inital ---> 20= (.25)v -0 since it starts from rest ----> velocity = 80 m/s

then i used the equation Vf^2 = V init ^2 + 2 a change y ----> 6400= 2(30.19) change y , since v init was 0 it cancels ----> and got change in y = 105.995 m

for D

i used the formula: delta y =( velocity init ) ( time) + 1/2 (acceleration ) (time^2) ---> 105.995= 1/2 (30.19) (time^2) since velocity init was 0 that part cancels ----> t= 2.65 s, and it would make sense because its after 2 s

Last edited: Oct 29, 2009
2. Oct 29, 2009

### jdwood983

Re: can some one just check my work plz

Why are you using impulse? This looks to be a straight-forward kinematics question, you should just use your Newtonian kinematics equations:

$$y=v_{y0}t+\frac{1}{2}a_yt^2$$

$$v_{yf}^2=v_{y0}^2+2a_y(y-y_0)$$

$$v_{yf}=v_{y0}+a_yt$$

3. Oct 29, 2009

### robair

Re: can some one just check my work plz

well because in the question it gives you impulse....

4. Oct 29, 2009

### jdwood983

Re: can some one just check my work plz

Right...seemed to have skipped over that word.

So

$$\mathbf{F}_{eng}\Delta t=\Delta p\rightarrow \mathbf{F}_{eng}=10 \mathrm{N}$$

I agree here. But I get

$$\sum\mathbf{F}=\mathbf{F}_{eng}-\mathbf{F}_{grav}=m\mathbf{a}\rightarrow \mathbf{a}=\frac{1}{m}\left(\mathbf{F}_{eng}-\mathbf{F}_{grav}\right)=\frac{1}{0.25}\left(10-9.8\right)=0.8 \mathrm{m/s^2}$$

5. Oct 31, 2009

### robair

Re: can some one just check my work plz

well i actually asked my physics teacher and he said i got A right so acceleration is 30.2 but i got B and C wrong...... he gave me the right answers for B (246m) and C (8.15s) so the thing right now is to figure out what i did wrong in B and C

6. Oct 31, 2009

### jdwood983

Re: can some one just check my work plz

Hmm....I get different answers than your teacher, particularly on the time part.

This is how I went about solving it:
The height of the rocket immediately after the engine burns out is given by

$$y_b=at^2=(30.2)(2)^2=120.8\,\mathrm{m}$$

The velocity at this point is given by

$$v_b=v_0+at=0+30.2\cdot2=60.4\,\mathrm{m/s}$$

So at the end of the engine burn, the rocket has total energy

$$E=K+U=\frac{1}{2}mv_b^2+mgy_b=752\,\mathrm{J}$$

At the maximum height, the particle has zero velocity but through the conservation of energy, it has the same energy:

$$E=U=mgy_{max}=752\rightarrow y_{max}=307\,\mathrm{m}$$

The total time of this is then given by

$$y_{max}=at^2\rightarrow t=\sqrt{\frac{y_{max}}{a}}= \sqrt{\frac{307}{30.2}}=3.2\,\mathrm{s}$$

Even using your teacher's result of 246 m for $y_{max}$ gives me a value of 2.85 s; however if you forget to take the square root of this number, you do get 8.15 (that is, if you say [tex]t=y_{max}/a=246/30.2=8.15[/itex], but this is not correct).