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					           Projectile motion is symmetrical around the
            “maximum height” of the projectile's path
The Acceleration is
always 9.8m/s2                                           SpeedA = SpeedI
Downward!!!!!                  E                         SpeedB = SpeedH
                           D                             SpeedC = SpeedG
                                      F                  SpeedD = SpeedF
                 C                               G

            B                                        H

    A
                                                           I

                 X    up              X   down
                        Xup = Xdown
Since Vx is constant and X up = X           From A to E speed slows down
                down                             From E to I speed speeds up
                Then
        Time up = Time down                              Speed is slowest at E
As a projectile reaches
maximum height the velocity     At max height Vy is ALWAYS 0 m/s!!
 slows down and flattens        The acceleration is, however,
out.                            9.8 m/s2 downward
Vy goes to 0 m/s at a rate of
9.8m/s2,
Vx is a constant




                As the projectile falls from
                maximum height the velocity increases
                and becomes more vertical
                Vy increases at a rate of 9.8m/s2
                VX is constant
   If a projectile’s path is symmetrical then The
    Time to reach maximum height equals one
    half of the total time of the projectiles
    motion.




   Knowing the time to reach maximum height we
    can use the position equation (the same equation
    we always used for projectile motion) to find the
    maximum height of the projectile.
 A soccer ball is kicked from ground level with a speed of 20 m/s
 and an angle of 30 degrees. What is the ball’s Maximum Height?



                                           (Half of range, Hmax)



      Vi = 20 m/s @
      30O                                                   (range, 0m)
                                 H   max

  (0m, 0m)

                               Range

Set up: 1st Get a Visual: Draw the situation
         2nd Break the problem into components:
                Draw X & Y axis's and label the initial and final points
                of the projectile. (You can have the origin anywhere)
                                             tup = 1.02 sec
                                           (Half of range, Hmax)
                                            (17.665,Hmax)


       Vi = 20 m/s @
       30O                                                    (35.33 m, 0m)
                                H   max

(0m, 0m)                                                         ttotal = 2.04 sec

                              Range

                     PosYf = PosYi + VYit +(½)ayt2
Hmax   = 0 m + (10 m/sec) (1.02 sec) +(1/2)(-9.8 m/sec2) (1.02 sec)     2

                              Hmax 5.1 m
   If the path is not symmetrical we can not
    simply divide the total time by two because
    tup = tdown
   In stead we need to look at the velocity of the
    projectile to find the time.
   We know that at maximum height Vy must be
    0 m/s.
   Since we are looking for time and we do not
    know the final position we can use the
    velocity equation to find time
   Vf = Vi + at
   But looking at the Y axis only
   Vfy = Viy + agt
   0 m/sec = Vi*Sinq + agtup
   So we can say that : tup = -(Vi*Sinq)/ag
   And now that we know tup we can use the
    position equation for the Y axis and find Hmax
    A soccer ball is kicked from ground level with a speed of 20 m/s
    and an angle of 30 degrees. What is the ball’s Maximum Height?
                                                (Half of range, Hmax)



         Vi = 20 m/s @
         30O                                                     (range, 0m)
                                   H   max

     (0m, 0m)

                                       Range

    Vfy = Viy + agt                            PosYf = PosYi + VYit +(½)ayt2
0 m/sec= 20m/sec*Sin30O + (-9.8 m/sec2)tup
           Hmax = 0 m + (10 m/sec) (1.02 sec) +(1/2)(-9.8 m/sec2) (1.02 sec)   2

tup = -(20m/sec*Sin30O )/ (-9.8 m/sec2)
                                                             Hmax 5.1 m
    tup = 1.02 sec
   Using both the velocity equation to find time
    up, and the position equation to find
    maximum height will ALWAYS work.
   However, it is a two step process. And nice
    would it be if there was a shorter way of
    doing this?
   Well, there is.
   Any time you have to use both the velocity
    and position equation together you can use
    the time independent equation instead.
   We start with the time independent formula
    ◦ Vf2 = Vi2 + 2a(Posf – Posi)
   We now apply it to the Y axis only
   Vyf2 = Vyi2 + 2ay(Posyf – Posyi)
   For Hmax we know
    ◦   Vyf= 0 m/s
    ◦   Vyi = ViSinq
    ◦   a =ag
    ◦   Posf= Hmax
    ◦   Posi = starting height {Hi} (This is usually 0 m)
   So we can know say
       (0 m/s)2 = (ViSinq)2 + 2(ag)(Hmax – Hi)
   This means that
    ◦ Hmax = [-(ViSinq)2/(2ag)] + Hi
A soccer ball is kicked from ground level with a speed of 20 m/s
and an angle of 30 degrees. What is the ball’s Maximum Height?
                                              (Half of range, Hmax)



    Vi = 20 m/s @
    30O                                                       (range, 0m)
                                  H   max

(0m, 0m)

                                      Range

                      Vyf2 = Vyi2 + 2ay(Posyf – Posyi)

           (0 m/s)2 = (20m/s*Sin 30)2 + 2(-9.8 m/s2)(Hmax – 0m)

                    Hmax= -(20m/s*Sin 30)2/ [2*(-9.8 m/s2)]

                                 Hmax 5.1 m

				
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