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 at 10m/s with no gravity looks like this…………….

Put them both together………

Regardless of the air resistance, the vertical and the horizontal components of velocity of an object in projectile motion are independent. Slowing down in +y dir. Speeding up in -y dir. Constant speed in +x dir. a x = 0 a y = -g Projectile Motion

When the magnitude of the velocity is given and its direction specified then its components can be computed easily x y V VXVX VYVY V X = Vcos   V Y = Vsin  You must use the polar angle in these formulas.

Example: Find the x and y components of the initial velocity vector shown if v = 10 and  = 60 degrees. v ix = v i cos  = 10 cos(60) = 5.00 m/s v iy = v i sin  = 10sin(60) = 8.66 m/s v ix = v i cos  = 10 cos(60) = 5.00 m/s v iy = v i sin  = 10sin(60) = 8.66 m/s V i = 10  = 60 o

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.

Projectile Motion Equations Vertical Direction  y = v iy t - 5t 2 v fy = v iy - 10t Vertical Direction  y = v iy t - 5t 2 v fy = v iy - 10t Horizontal Direction  x = v ix t v fx = v ix Horizontal Direction  x = v ix t v fx = v ix Note: We are using g = 10 Auxiliary Equations v ix = v i cos  v iy = v i sin  Auxiliary Equations v ix = v i cos  v iy = v i sin 

Case 1 : Fired Horizontally

Case 1 : Example A cannon is fired horizontally at 50m/s from a cliff 20m tall. 1.How long is it in the air? 2.What is the horizontal range.

Case 2 :Ground-to-Ground Motion Time of Flight, T = ( 2 v i sin  /g Range, R= v i 2 sin (2  /  g Maximum Height H, H = (v i sin  ) 2 /2g

Case 2: 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? 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= v i 2 sin (2  =  563) 2 sin(120)/10 = 27450m = 27km ( Due to air resistance, the actual range is less than this.) Range, R= v i 2 sin (2  =  563) 2 sin(120)/10 = 27450m = 27km ( Due to air resistance, the actual range is less than this.) Time of Flight, T = (2v i sinq) /g = 2(563) sin(60)/10 = 97.5 seconds Time of Flight, T = (2v i sinq) /g = 2(563) sin(60)/10 = 97.5 seconds

The Trajectory Equation (1) y f = v i (sin  ) – (g/2)t 2 ( 2) x f = v i (cos  )t To get the path or trajectory equation for a projectile, we need to find y as a function of x. Starting at (0,0): If you solve (2) for t and plug it into (1) you get:

Trajectory This has the form y = ax +bx 2, a parabola.

Example: Find the equation of the parabola if v = 10 and  = 60 degrees.

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