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Find the dx for an object projected at 100 m/s from 20 degrees to 70 degrees in 5 degree intervals. How would you exactly do this? Would you use the equation Ay = A sin X?
Skotster
Dec6-04, 05:57 PM
break up your Vi=100m/s at 20 degrees into the x&y components
Vix=Vi*cos(20 degrees)
Viy=Vi*sin(20 degrees)
assuming you are on earth and ingoring air friction
use
Vf=Vi+(-9.8m/s^2)*t
to get
t=(Vf-Vi)/(-9.8m/s^2)
lettin Vf=0 to find time to max height
double it to get the total air time
use total time in
dx=Vix*t
Do I have to do Viy=Vi*sin(X) repeatedly until the equation is Viy=Vi*sin(70)? Do you have a screen name so we can talk one-on-one?
break up your Vi=100m/s at 20 degrees into the x&y components
Vix=Vi*cos(20 degrees)
Viy=Vi*sin(20 degrees)
assuming you are on earth and ingoring air friction
use
Vf=Vi+(-9.8m/s^2)*t
to get
t=(Vf-Vi)/(-9.8m/s^2)
lettin Vf=0 to find time to max height
double it to get the total air time
use total time in
dx=Vix*t
Skotster
Dec6-04, 06:38 PM
you could use theta in terms of n like this
theta=20+5n
where n is a whole number between 0 and 10
use your basic algebra to substitute in variables, ie: the for t in dx=Vix*t put dx=Vix*-2Viy/(-9.8m/s^2)
do the same for Vix and Viy, and then again for theta.
SN: PBGartist (aim)
58362144 (icq)
pbgartist@hotmail.com (msn)
Real_Skotster (yim)
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