The Kinematics of Projectile Motion

Report
Kinematic Equations and
Projectile Motion
By Corina Bot
Projectile Motion
The speed in the x-direction is
constant; in the y-direction
the object moves with
constant acceleration g.
This photograph shows two balls
that start to fall at the same time.
The one on the right has an initial
speed in the x-direction. It can be
seen that vertical positions of the
two balls are identical at identical
times, while the horizontal
position of the yellow ball
increases linearly.
wikipedia.org
- Package follows a parabolic path and remains
directly below the plane at all times.
- As the package falls, it undergoes a vertical
acceleration;  there is a change in its vertical
velocity.
- This vertical acceleration is attributed to the
downward force of gravity which acts upon the
package.
http://www.physicsclassroom.com/
Kinematic Equations for Constant Acceleration in Two Dimensions
x Component (horizontal)
vx = vxo + axt
x = xo + vxot + ½ axt2
vx = vxo + 2ax(x – xo)
y Component (vertical)
vy = vyo + ayt
y = yo + vyot + ½ ayt2
vy = vyo + 2ay(y – yo)
- A cannonball being launched at an angle
from a cannon atop of a very high cliff.
- The cannonball follows a parabolic path.
- As the cannonball rises towards its peak, it
undergoes a downward acceleration. An
upwardly moving cannonball which is
slowing down is said to be undergoing a
downward acceleration.
v0y
q0
v0x
http://www.physicsclassroom.com/
Conceptual Example 3-6/pag. 59: Where does the apple
land?
A child sits upright in a wagon
which is moving to the right at
constant speed as shown. The
child extends her hand and
throws an apple straight upward
(from her own point of view),
while the wagon continues to
travel forward at constant
speed. If air resistance is
neglected, will the apple land (a)
behind the wagon, (b) in the
wagon, or (c) in front of the
wagon?
"Physics" by Giancoli, 6th ed.
A child sits upright in a wagon which is
moving to the right at constant speed
v0=20m/s, as indicated. The child extends
her hand and throws an apple at an angle
a = 27⁰, while the wagon continues to
travel forward at constant speed. If air
resistance is neglected, calculate distance
traveled by the cart until apple reaches
back the child’s hand.
v0=20m/s
General equations
a = 27⁰
d?
 x  v0 x  t


1
2
y

v

t

a

t
0
y

2

y
v0y
Particular equations
x  d  v ox  t  v o cos a   t
y  0  v oy  t 
1
2
g t
2
v0
time = t
a
v0x
x
d=?


1
2 
 v o sin a   t 
g t 
2

t=?
d=?
Particular equations
x  d  v ox  t  v o cos a   t
y  0  v oy  t 
1
g t
2
2


1
2
 v o sin a   t  g  t 
2

v0=20m/s
t=?
d=?
a = 27⁰
d?
y
From horizontal (x) equation:
v0y
x  d  v ox  t  v o cos a   t
d  v o cos a   t
m
t
(1)
Substitute t in equation (3) to get d:
m
t
s
d  17 . 82
m
s
d  32 . 9 m
 1 .8 s
d=?
x
y  0  v oy  t 
s
d  17 . 82
a
v0x
From vertical (y) equation:
m


d   20
 cos 27   t
s


d  17 . 82
v0
time = t
v o sin a   t 
1
1
g t
2
g t
2
2
  v o sin a   t 
 0
2
m
1
m 2


 sin 27   t   9 . 8 2  t  0
 20
s
2
s


9 . 07  t  4 . 9  t
2
9 . 07  4 . 9  t  0
4 . 9  t  9 . 07
t  1 . 85 s
 0
:t
1
2
g t
2

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