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2D Motion Question

  • Thread starter JWest
  • Start date
  • #1
12
0
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?
 

Answers and Replies

  • #2
8
0
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
 
  • #3
12
0
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?





Skotster said:
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
 
  • #4
8
0
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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