Trouble With Projectile Motion (Can someone check my workings please?)

[skeletalclown]

Hi everyone, this is my first post here, and it's about our projectile motion extended investigation.

Basically, we fired air powered rockets at various angles to see if the optimal angle for horizo0ntal distance is actually 45 degrees, even despite air resistance.

At first I measured the time taken to fall vertically (slightly greater than at angles, presumably due to air resistance i just averaged out all the times), and measured 4.6 seconds

The distances obtained were in the order of 60-120m, (my books not here right now), with 30 and 45 degrees being the farthest.
Anyway, my actual question, we have to calculate the THEORETICAL horizontal distance traveled, and i'm having trouble:

t=4.6s a=-10ms^-2
First i tried to find the vertical initial velocity: v=u+at (v=0 at zenith)
0=u-46
u=46ms^-1
46 meters per second seemed WAY too high, and things get even weirder when i try to figure out the horizontal velocity:

A=O/cos(theta), gave the results:

30 degrees= 53.1ms^-1
45 degrees= 65.1ms^-1
60 degrees= 92ms^-1
75 degrees= 177.7ms^-1

Not only is 45 not the largest, but they seem WAY too large, at this rate, using x=vt, 45 would be the second SHORTEST distance, and the 75 DEGREE rocket would travel nearly a kilometer,

Im not very good at physics, and ive been researching and trying this for hours, ANY help would be really appreciated.

[skeletalclown]

Sorry everyone, i wrote this very late at night and have since realised my mistakes. For some reason my internet cuts out whenever i try to edit the original post, so i'll just place my new calculations here:

I would REALLY appreciate if somebody would go over this and just check to make sure im on the right track:

Info: 30 degrees=110m, 3.5s. 45 degrees=110m, 4.5s. 60 degrees=60m, 4.5s

To get the vertical component of initial velocity:

v=u+at            30 degrees=17.5ms^-1   45 degrees=23ms^-1         60 degrees=22.5ms^-1

Is this step okay???

Then to find the horizontal component: O/tan(angle)=a           

30 degrees=30.3m   45 degrees=23m         60 degrees=13m

To find theoretical distance (the entire point), X=vt

30 degrees=109m  45 degrees=105.8         60 degrees=59m

These results are very close to those actually measured, but I'm a little concerned about my workings and would appreciate it if somebody could look over them??? I'm also worried about 30 degrees giving better results than 35, but i assume this is due to the larger amount of surface area available to the air resistance at the larger angle of 45 degrees, would i be right there?? Thanks in advance guys!

[Thorium]

I checked your calculations in your second post and they looked correct. And you shouldnt expect your experiment to be perfect with no errors, primarily due to the fact that we do not live in an ideal world. So this leaves some gap for discussion in your  practical report. And you have already mentioned a source of error, which is the air resistance. Other sources of error might be reaction time for pressing the stopwatch button if you used one, errors due to slight estimations in determining the distance travelled etc

Hope that helps

[skeletalclown]

Thank you!! its very reassuring knowing i havent totally stuffed up yeah, a very large section of the prac report is going to be on errors i think (20 METRES DISPLACEMENT FOR THE 90 DEGREE LAUNCH DUE TO WIND ALONE), and yeah, thanks again!!