Projectile Motion
•
The Horizontal Motion
•
The Vertical Motion
•
The Trajectory
•
The Horizontal Range
•
Example
•
Homework
1
The Horizontal Motion
ax = 0
⇒
vxf = vxi = vi cos θi = constant
xf = xi + vxit = xi + (vi cos θi) t
2
(1)
The Vertical Motion
ay = −g
1
1
yf = yi + vyit − gt2 = yi + (vi sin θi) t − gt2
2
2
vyf = vyi − gt = vi sin θi − gt
(2)
(3)
2
2
vyf
= vyi
− 2g (yf − yi) = (vi sin θi)2 − 2g (yf − yi) (4)
3
The Trajectory
•
•
Solving eqn. (1) for t we have
xf − x i
t=
vi cos θi
Substituting this into eqn. (2) and setting xi = yi =
0,we get
2
1
x
x
f
f


 −

g 
yf = (vi sin θi) 
vi cos θi
2 vi cos θi



yf = (tan θi) xf − 

•


g

2

x

2 (vi cos θi)2 f

(5)
Eqn. (5) has the form y = ax + bx2 where a and
b are constants ⇒ the trajectory of a projectile is
parabolic
4
The Horizontal Range
xf − x i = R
Eqn. (1)
Eqn. (2)
y f − yi = 0
R = (vi cos θi) t
1
⇒ 0 = (vi sin θi) t − gt2
2
2vi sin θi
t=
g
⇒
5
The Horizontal Range (cont’d)
2vi2
R=
sin θi cos θi
g
sin 2θi = 2 sin θi cos θi
vi2
R = sin 2θi
g
Note: R is a maximum when sin 2θi = 1 ⇒ 2θi = 90◦ or
θi = 45◦
6
Example
A soccer player kicks a ball at an angle of 38◦ from the
horizontal, with an initial speed of 15 m/s. (a) How long
is the ball in the air? (b) How far down the field does the
ball land? (c) How high does the ball go?
7
Example Solution
A soccer player kicks a ball at an angle of 38◦ from the
horizontal, with an initial speed of 15 m/s. (a) How long
is the ball in the air?
vi = 15 m/s
θi = 38◦
yf = y i
1 2
yf − yi = (vi sin θi) t − gt = 0
2
2vi sin θi 2 (15 m/s) sin 38◦
t=
=
= 1.9 s
g
9.8 m/s2
(b) How far down the field does the ball land?
xf −xi = (vi cos θi) t = (15 m/s) (cos 38◦) (1.9 s) = 22 m
(c) How high does the ball go?
vyf = 0
2
= (vi sin θi)2 − 2g (yf − yi) = 0
vyf
vi2 sin2 θi (15 m/s) sin2 38◦
=
= 4.4 m
yf − y i =
2g
2 (9.8 m/s2)
8
Homework Set 7 - Due Fri. Sept. 24
•
Read Section 3.3
•
Do Problems 3.9, 3.11 & 3.14
9