graph for projectile... (1 Viewer)

[Belle]

Originally posted by ND
If you drew a line through (0, 0) and (.6, 1.85), then you drew a line with gradient t.

I know, but Inhuman was right, I was talking about the first bit with the velocity I went the complicated way and didn't think of the simple way and didn't even end up with the right answer.

Oh well. My graph will be wrong but maybe I can get a mark for the pretty scale I drew or something... or drawing a straight line...?

 

[Takuya]

Originally posted by Bannanafish
er you had to plot it for different velocities

I used 1m/s, then 1.85m/s then 3m/s.

So effectively I had 3 different flat lines of different lengths on the page...

 

[Squirrel]

my stupid graph had a negative gradient.

i knew it was wrong but i did it anyway.

 

[Bannanafish]

i mean plot a continuous set of velocities, so one straight line y=mx

 

[Takuya]

I wonder if some people read what they were actually plotting...

HORIZONTAL VELOCITY VS RANGE.

Horizontal velocity (y-axis) is CONSTANT!!! Therefore it's a straight, flat line!

 

[Constip8edSkunk]

u miss read the question, it says graph it when the velocity is varied...

even by assuming velocity is constant, u can never get a graph because the range would be constant too, unless you have a adjustable table

 

[walla]

The time taken is constant (unaffected by horizontal velocity).

x= ut

so the range is directly proportional to horizontal velocity - straight line graph.

 

[Sprach]

so the graph looked like this?

| .
| .
| .
| .
|._________
lol join the dots
is that what it looked like? a straight line graph?

 

[Sprach]

rofl that didn't work, it was supposed to be y = x lol

 

[mikehunt]

i got

l .
l .
l .
l .
l.______

 

[jims]

not sure, but i used dx = u(x)t so as u(x) increases, dx increases linearly.

 

[dnt stop runnin]

is the graph a straight line or wat???

 

[+Po1ntDeXt3r+]

yer its straight...mine was m=sqrt(2/g) i think it may have been m=sqrt(g/2)
i stuffed it up TWICE
and my points was (0,0) and (1.85,0.6) cos i thought the range was dependent on the velocity. was i wrong?

 

[Amish]

I had a good 10mins at the end to think solidly about this question...

I came to the conclsion that it was a linear graph that went through the point (1.85, 0.6). As they asked to show at different velocities my horizontal axis was say....horizontal velocity and vertical was range. I used 6 x values being 1,1.85, 2, 3, 4, 5 and then connected them using straight line so it was pretty much a linear graph with equation y = (12/37)x

Hopefully i got it right

 

[iH]

it was a straight line.

as velocity increases so does range.

t is constant because it is a variable affected only by gravity. assuming gravity and the table height are the same, t stays the same.

one straight line y = x

goes through the origin

and if you were smart you also plotted the point correctly as (0.6, 1.875) <--- not sure if that's the exact number (the y value) cos it was hours ago

 

[marky]

my friends said it was jst a straight line since bvelocity in the y direction is constant????

 

[smeyo]

yeah a straight line with gradient =t going through (0,0) and (1.85,0.6)

 

[Sprach]

.

 

[dnt stop runnin]

i did a straight line but i didnt start it at the origin....u rekon ill get a mark??

 

[iH]

.





you might get a mark but with all the wholistic marking. who knows...it kinda shows that u weren't exactly sure what u were talking about....cos the entire thing is determined to figure out if u know that s = vt....and if u knew that you'd see that when v = 0, so is s

Sam