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View Full Version : (Projectile Motion)How to derive the equation??


herfox
Apr17-05, 07:09 PM
I made a device which use elastic band to launch a ping pong ball.
I have measured the distance travelled with different length of stretch at constant 36 degree launch angle. The graph plotted out is kind of curved, not straight line. And now I am asked to derive an equation to determine distance as a function of stretch....
Does anyone know how to do this? Cuz I spent almost a day still duno how to do it.

christinono
Apr17-05, 07:12 PM
I made a device which use elastic band to launch a ping pong ball.
I have measured the distance travelled with different length of stretch at constant 36 degree launch angle. The graph plotted out is kind of curved, not straight line. And now I am asked to derive an equation to determine distance as a function of stretch....
Does anyone know how to do this? Cuz I spent almost a day still duno how to do it.
First, determine what kind of relationship it is (i.e. exponential, root...). WHat kind is it?

herfox
Apr17-05, 07:17 PM
Well... i dont really know actually. I guess its root maybe?
heres the data
Stretch(cm) Distance(m)
10.7 5.684
15.35 7.955
20.12 9.98
24.96 10.365
29.83 11.725

christinono
Apr17-05, 07:22 PM
Well... i dont really know actually. I guess its root maybe?
heres the data
Stretch(cm) Distance(m)
10.7 5.684
15.35 7.955
20.12 9.98
24.96 10.365
29.83 11.725
Wait a minute, let me enter the data in my calc

christinono
Apr17-05, 07:26 PM
Well, it's hard to tell for sure. If you had more points, it would be easier to determine. The first 3 points look linear, but with the last 2 points, it looks more like a root relationship.

herfox
Apr17-05, 07:28 PM
oh.... yaa that's what the graph looks like...
alright maybe I gotta do more points then.
Anywayz, Thanks a lot!

OlderDan
Apr17-05, 07:38 PM
If you think the data is related by a power function you can try a power regression on your calculator. In the old days, the power was usually found by graphing the log of the distance and the log of the stretch to decipher the exponent from the slope of what would then be a linear graph. If your calculator can't do it, I can elaborate on the log-log approach

futb0l
Apr18-05, 03:01 AM
You may want to try.. y/x and y/x^2 ... y/sqrt(x) and so on, which gets you the constant of proportionality k ... and see which one is more consistent

kusal
Apr19-05, 11:27 PM
of course it's a curve
here's the equation

v=u+at (vertical motion)
-u sin36=u sin36 -gt
t=2u sin36/g

s=ut
s= u sin36 * 2 u sin36/g
s= u*u *2 *sin36 *cos36 / g

but
mv*v/2= kx*x/2
mu*u/2=kx*x/2

u*u=kx*x/m

then

s=kx*x *2 *sin36 *cos36 / gm

as all the other variables remain constant

if a=k*2 *sin36 *cos36 / gm

s= a * x^2