# Level Projectile PPT by wuzhengqin

VIEWS: 2 PAGES: 16

• pg 1
```									Physics Honors
Good News/Bad News: These are the same formulas we
used for linear motion. Do you know them?

  o  at
d vt                       1
2
2

vf v at
o

v 2
2         2
f            o
If the answer is “NO”, then memorize them NOW!
 We assume NO AIR RESISTANCE! (Welcome to “Newtonia”),
therefore…
 The path of a projectile is a parabola.
 Horizontal motion is constant velocity.
v
x constant

a
x 0
 Vertical motion is in “free-fall”.
ay  9.81 m/s2
 Vertical velocity at the top of the path is zero
vy top  0
 Time is the same for both horizontal and vertical motions.
horizontal or “x” – direction                 vertical or “y” – direction
0

 x v t  a t
d    ox
1
2 x
2               v t  a t
d y       oy
1
2 y
2

vfx v a t                                 vfy v ayt
0

ox x                                        oy

v v 2 x x v v 2 y y
0          2         2
2          2
fx   a d    ox
a d                      fy         oy
Remember that for projectiles, the horizontal and vertical motions must be
separated and analyzed independently. Remember that “ax” is zero and “ay” is
acceleration due to gravity “a = - 9.81 m/s2”.
A cannon ball is shot horizontally from a cliff.

vx
What do we know?
For all projectiles…
v
x constant

a
x 0
Δdy
.
a  9 m/s
y   81 2
v topv y 0
y     o

Range,   Δdx                    t tx ty

Hint: You should always list your known values at the beginning of any
problem and assign those values variables.
A cannon ball is shot horizontally from a cliff.

vx
Knowns:
v
x constan

a
x 0
.
a  9 m/s
y    81 2
Δdy                                                             o
v top v y 0
y

t x  y
t t

Range, Δdx
Remember to keep the horizontal and vertical motions separate. Time
is the factor that ties them together. Since the vertical motion is treated
that to find time first.
A cannon ball is shot horizontally at 5 m/s from a cliff that is 35 m high.
Find time, t = ? Find range, dx = ? Find final velocity, vf = ?
Vx= 5 m/s                                         Knowns:
Givens:                v
x constan

Δdy= -35 m                                                   a
x 0
v 5m/s
x
.
a  9 m/s
y    81 2
 y  m v v y 0
d    35  ytop o

t x  y
t t

Range, Δ dx
Add the given values to our list of known values. Now that the diagram
is drawn and labeled and we have identified and listed all of our
“known” and “given” values for the problem, let’s begin by finding time.
A cannon ball is shot horizontally at 5 m/s from a cliff that is 35 m high.
Find time of flight, t = ? Find range, dx = ? Find final velocity, vf = ?
Vx= 5 m/s                     Givens:             Knowns:

Δdy= -35 m
v  m/s v 
x 5    x constan

 y  ma
d    35 x 0
.
a  9 m/s
y    81 2
o
v top v y 0
y

t x  y
t t
Range, Δdx
Since we know more values for vertical motion, let’s use it to find time.
Now   .          t
4 2
d y  vo y t  2    1 at 2
solve
35 905
for t… t        35                67.
2 se
 35  0        1
2   ( 9.81)t  2
905
4.
A cannon ball is shot horizontally at 5 m/s from a cliff that is 35 m
high. Find time of flight, t = ? Find range, Δdx = ? Find final
velocity, vf = ? V = 5 m/s                                         Knowns:
x                          Givens:

Δdy= -35 m
v  m/s v 
x 5    x constan

 y  ma
d    35 x 0
.
a  9 m/s
y    81 2
Calculated:
o
v top v y 0
t  2.67s          y

t x  y
t t
Range, Δdx
Now that we know time, let’s find dx. Remember that horizontal motion is
constant. Let’s use the distance formula again. This time in the x – direction…

 t
 v 2x
d it a
x x
1 0     2

d 5  s 13
2 
   67. m
x   . 
m/s 36
A cannon ball is shot horizontally at 5 m/s from a cliff that is 35 m high. Find
time of flight, t = ? Find range, Δdx = ? Find final velocity, vf = ?
Vx= 5 m/s                    Givens:                Knowns:

Δdy= -35 m
v  m/s v 
x 5    x constan

Vf           y  ma
d    35 x 0
.
a  9 m/s
y    81 2
Calculated:
o
v top v y 0
t  2.67s              y

Range, Δdx                       d x  13.36 m        t x  y
t t

Final velocity requires a little more thought. Remember that velocity is a vector
quantity so we must state our answer as a magnitude (speed that the projectile
strikes the ground) and direction (angle the projectile strikes the ground). Also
remember horizontal velocity is constant, therefore the projectile will never
strike the ground exactly at 90°. That means we need to look at the horizontal
(x) and vertical (y) components that make up the final velocity.
A cannon ball is shot horizontally at 5 m/s from a cliff that is 35 m high. Find
time of flight, t = ? Find range, Δdx = ? Find final velocity, vf = ?
Vx= 5 m/s                     Givens:               Knowns:

Δdy= -35 m
v  m/s v 
x  5    x constan

Vf         y  ma
d    35 x 0   2
y .
a  9 m/s
81
Calculated:
o
v top v y 0
t  2.67s              y

Range, Δ dx            θ         x  . m  x  y
d 13  36 t t t
Let’s look more closely
at the vector, vf. To                                    So, we have the x-
help see it better, let’s                                component already due to
exaggerate the angle.                         Vf         the fact that horizontal
Vfy
Since x- and y- motion                                   velocity is constant. Before
are separate, there                           θ          we can find vf, we must find
must be components.                  Vfx = 5 m/s         the vertical component, vfy.
A cannon ball is shot horizontally at 5 m/s from a cliff that is 35 m high. Find
time of flight, t = ? Find range, Δdx = ? Find final velocity, vf = ?
Vx= 5 m/s                    Givens:                Knowns:

Δdy= -35 m
v  m/s v 
x 5    x constan

Vf                   y  ma
d    35 x 0
v y v y 
f    o   at                           .
a  9 m/s
y    81 2
Calculated:
o
v top v y 0
t  2.67s            y

Range,Δdx           θ                   x  . m  x  y
d 13  36 t t t
To find vfy, remember                                         Calculating vfy:
that vertical motion is in
vy  o 
vy at
“free-fall” so it is                                             f
Vf
accelerated by gravity             Vfy                          vy  . m/s. s
f  0 (981 2 2 )
)(67
from zero to some
value just before it hits                     θ                      . m/s
vy  26
f    2
the ground.                          Vfx = 5 m/s                  Still not finished. Gotta put
components together for final.
A cannon ball is shot horizontally at 5 m/s from a cliff that is 35 m high. Find
time of flight, t = ? Find range, Δdx = ? Find final velocity, vf = ?
Vx= 5 m/s                    Givens:                Knowns:

Δdy= -35 m
v  m/s v 
x 5    x constan

Vf                   y  ma
d    35 x 0
v y v y 
f    o   at                          .
a  9 m/s
y    81 2
Calculated:
o
v top v y 0
t  2.67s           y

Range,Δdx         θ                   x  . m  x  y
d 13  36 t t t
Now we know both the x-                                     Putting it together to calculate vf:
and y- components of the
 .2
v  v fy 5 26
22      2
final velocity vector. We                                  f   x  v      2
need to put them together                    Vf
26 
v  . m/s
f   7      (spe
magnitu
for magnitude and
direction of final velocity.                θ               tan2 79
 5
  26  .  direc
1 .
2
Vfx = 5 m/s
Final Velocity = 26.7m/s@280.2°
Warm up (pg.2)
►A  rock is kicked horizontally off of a 457 m
tall cliff.
A.) How long does it take to hit the ground?
B.) how far does the rock land from the base
of the cliff if the horizontal velocity is
15 m/s
Projectile Lab
►   At your station you have a paper ball, meter stick, and stop watch.
►   Your job is to throw the paper ball with horizontal projection (don’t
throw it hard).
-Record the amount of time it takes for the ball to hit the ground
and measure the distance the ball travels in the air.
- Calculate the horizontal velocity of the ball.
►   Calculate the height at which the ball was released using the time you
recorded. Now compare your calculated value to the actual value (you
will have to measure this).
►   Using the actual height of the release point of the ball, calculate the
amount of time the ball should have been in the air.
►   Using the time you just calculated and the actual release point of the
ball (what you measured), calculate the vertical velocity of the ball just
before it hits the ground

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