# Slide Katy ISD Range

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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
◦ 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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 views: 9 posted: 12/28/2010 language: English pages: 12
Description: Slide Katy ISD Range