1 / 24

Projectile Motion-Starter

Projectile Motion-Starter. What is the path that the bike and the water take called?. Projectile Motion. A ball dropped from rest looks like this……. Projectile Motion. A ball rolled off the table with no gravity looks like this……………. Put them both together………. Another View.

siusan
Download Presentation

Projectile Motion-Starter

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Projectile Motion-Starter What is the path that the bike and the water take called?

  2. Projectile Motion A ball dropped from rest looks like this……..

  3. Projectile Motion A ball rolled off the table with no gravity looks like this…………….

  4. Put them both together………

  5. Another View The projectile falls away from a straight line it would have taken if there were no gravity by the same distance it falls from rest.

  6. A vector v in the x-y plane can be represented by its perpendicular components called vx and vy. y Components vXand vY can be positive, negative, or zero. The quadrant that vector A lies in dictates the sign of the components. Components are scalars. V VY x VX

  7. When the magnitude of vector v is given and its direction specified then its componentscan be computed easily y VX= Vcosq V VY VY= Vsinq  x VX You must use the polar angle in these formulas.

  8. Example: Find the x and y components of the initial velocity vector shown if v = 10 and q = 60 degrees. Vi = 10 q = 60o vix = vi cosq = 10 cos(60) = 5.00 m/s viy= vi sinq = 10sin(60) = 8.66 m/s

  9. Projectile Motion Initial Conditions

  10. Projectile Motion Equations Vertical Direction yf = yi +viyt + (1/2)ayt2 yf = yi + (t/2) (viy+ vfy ) vfy = viy +ayt vfy2 = viy2+2ay (yf - yi ) (ay = -9.8m/s2) Horizontal Direction xf = xi +vixt vfx= vix Auxiliary Equations vix = vi cosq viy = vi sinq

  11. How To Use Them 1st: Calculate viy , and vix if required. vix = vi cosq viy = vi sinq 2nd: List all the variables: yf = xf = yi = xi = viy = vfx= vix = vfy = ay= -9.8 t = 3rd: Read the problem and fill in all you can. 4th: Pick an equation with just one of your unknowns.

  12. Example: 1st: Calculate viy , and vix if required. vix = vi cosq = 8cos(0) = 8 m/s viy = vi sinq =8sin(0) = 0 2nd: List all the variables: yf = 0 xf = ? yi = 80m xi = 0 viy = 0 vfx= vix = 8.0 m/s vfy = ? ay = -9.8 t = ? We need xf so we must use xf = xi +vixt But we don’t know t. So we start with an equation that has t, but not vfy . yf = yi +viyt + 1/2ayt2 0 = 80 -4.9t2 ; t = 4.04 seconds Then xf = vixt =8(4.04) = 32.3m

  13. Example: 2nd: List all the variables: yf = 0 xf = ? yi = 0 xi = 0 viy = 17.3 m/s vfx= vix = 10.0 m/s vfy = ? ay = -9.8 t = ? 1st: Calculate viy , and vix if required. vix = vi cosq = 20cos(60) = 10 m/s viy = vi sinq =20sin(60) = 17.3 m/s

  14. Continued…. 1st, let’s get t and xf . We start with an equation that has t, but not vfy . yf = yi +viyt + 1/2ayt2 0 = 0 + 17.3t -4.9t2 Dividing by t gives: 0 = 0 + 17.3 -4.9t t = 17.3/4.9 = 3.53 s Then xf = vixt =10(3.53) = 35.3m

  15. Continued…. To get the maximum height, we have to revisit the initial conditions, because this is a new problem. The final y position is now unknown and the final y-velocity is now zero at the top of the flight. List all the variables: yf = ? xf = ? yi = 0 xi = 0 viy = 17.3 m/s vfx= vix = 10.0 m/s vfy = 0 ay = -9.8 t = ? We need an equation with yf But without t, so we use vfy2 = viy2+2ay (yf - yi ) or 0 = 102 -19.6 yf Then yf = 100/19.6 = 5.10m

  16. Ground-to-Ground Motion:A Special Case Range (R) and Time of Flight (T)

  17. Ground-to-Ground Motion:A Special Case List all the variables: yf = 0 xf = R yi = 0 xi = 0 viy = vi sinq vfx= vix = vi cosq ay = -9.8 t = T yf = yi +viyt + 1/2ayt2 or 0 = 0 + vi sinq T + 1/2ayT2 or T = 2 vi sinq / (-ay) ( Using g = 9.8 = -ay ) then T = (2vi sinq) /g R = vixT = vi cosq (2 vi sinq ) / g , so R = vi 2sin (2q)/g

  18. Maximum Height, H vfy2 = viy2+2ay (yf - yi) 0 = (vi sinq )2 -2gH H = (vi sinq )2/2g H = (vi sinq )2 /2g

  19. Ground-to-Ground Motion:A Special Case Range , R=vi 2sin (2q)/g Time of Flight, T = (2vi sinq) /g Maximum Height H, H = (vi sinq )2 /2g

  20. Example The muzzle velocity of a Howitzer is 563 m/s. If it is elevated at 60 degrees, what is its range and time of flight? Range , R=vi 2sin (2q) = (563)2sin(120)/9.8 = 28010m = 28km ( Due to air resistance, the actual range is less than this.) Time of Flight, T = (2vi sinq) /g = 2(563) sin(60)/9.8 = 99.5 seconds

  21. The Trajectory Equation To get the path or trajectory equation for a projectile, we need to find y as a function of x. Starting at (0,0): (1) yf= vi(sinq) – (g/2)t2 (2) xf= vi(cosq )t If you solve (2) for t and plug it into (1) you get:

  22. Trajectory This has the form y = ax +bx2 , a parabola.

  23. Example: Find the equation of the parabola if v = 10 and q = 60 degrees. 2

  24. Exit The path of a projectile is given by: y = 2x –x2 What is the horizontal range? Hint: for what values of x is y = 0?

More Related