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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.

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Projectile Motion-Starter

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  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?

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