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Lecture Notes on Projectile Motion - Elements of Physics | PHYS 151, Study notes of Physics

Material Type: Notes; Class: Elements of Physics; Subject: Physics ; University: University of Nebraska - Lincoln; Term: Unknown 1989;

Typology: Study notes

Pre 2010

Uploaded on 08/30/2009

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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
== = =
== =
°
pf3

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PHYSICS 151 – Notes for Online Lecture

Projectile Motion

In this lecture we will look at projectile problems with a general launch angle – projectile launched with velocity v 0 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 v (^) 0x = v0 cosθ

The y component of velocity is v (^) 0y = v 0 sinθ

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 2

( .4 50 s) = 2.25 s, and at that time v (^) y = 0. So considering the motion from that time on and since v (^) y = v (^) 0 ygt ,

v (^) 0 y = F9 81 2 25 22 07 HG^

I KJ^

. m (. ) =. s

s m (^2) s

So, v 0 22 07 63 0

= 24 8 °

= . sin.

..

m s (^) m s

0 0 ms 0

1.50 m (^) 1.20m 1.25 s s

1.57 m s cos cos 40.

x x x

v x v t v v θ

= = = =

= = = °

Ex 9-3 : On a hot summer day, a young girl swings on a rope above the local swimming hole. When she lets go of the rope her initial velocity is 2.25 m/s at an angle of 35.0° above the horizontal. If she is in flight for 1.60 s, how high above the water was she when she let go of the rope?

Known: v0y = 1.29 m/s Solve: y NI: v (^) y

a = -9.81 m/s 2 t = 1.60 s

What was the girl’s greatest height above the water?

2 2 0 2 2 0

2

y y

y

v v ay

m v (^) s y m a m s

− ⎛^ ⎞

− ⎜^ ⎟

= = ⎝^ ⎠ =

So adding this to 10.5 m gives only 10.6 m.

( )

0 sin

2.25 sin 35

v oy v m s m s

= θ

= ⎛^ ⎞ ° ⎜ ⎟ ⎝ ⎠

=

You Try It!

( ) ( )

2

2 2

y v t oy at

m m s s s s m