PHYSICS 151 – Notes for Online Lecture #9
Projectile Motion
In this lecture we will look at projectile problems with a general
launch angle – projectile launched with velocity v0 at angle θ.
We will one again make use of the fact that the horizontal and
vertical motions are entirely independent. If we break the initial
velocity into horizontal and vertical components we can treat the
two motions separately with different initial velocities.
The x component of velocity is v0x = v0cosθ
The y component of velocity is v0y = v0sinθ
Ex. 9-1 A cork shoots out of a champagne bottle at an angle of 40.0° above the horizontal. If the cork
travels a horizontal distance of 1.50 m in 1.25 s, what was its initial speed?
We can ignore the vertical motion and just consider the horizontal motion as due to the horizontal
component of the initial velocity.
Ex. 9-2 The “hang time” of a punt is measured to be 4.50s. If the ball was kicked at an angle of 63.0°
above the horizontal and was caught at the same level from which it was kicked, what was its initial
speed?
The maximum height is achieved at time t=1
2450( . s) = 2.25 s,
and at that time vy=0.
So considering the motion from that time on and since vv
g
t
yy
=
−
0,
vy09 81 2 25 22 07=
F
H
GI
K
J=.(.).
m
ssm
s
2
So, v022 07
630 24 8=°=
.
sin . ..
m
sms
0
m
0s
0
1.50 m m
1.20
1.25 s s
1.20 1.57 m s
cos cos40.0
xx
x
x
vv
t
v
v
θ
== = =
== =
°
