PHYSICS 151 – Notes for Online Lecture 1.7
In the real world, we don’t exist in just one dimension ⇒ We need to describe how things move in two
and three dimensions! We will use subscripts for the velocities and accelerations to make it clear when
we’re referring to the x direction and when we’re referring to the y-direction.
x
y
ax = 0
v y = v y0 + a y t
x = vx t
1
y = v y 0t + a y t 2
2
v y 2 = v 2y0 + 2a y ( y − y 0 )
y = v yt =
(v
oy
+ vy )
2
t
Ex. The motion of a ball in flight
•
The path is parabolic in nature. You may remember studying things called conic sections in
geometry. You need special equations to describe complicated shapes like this. Luckily for us, it turns
out that motion in the x and y directions are independent of each other. That is, if I were to drop one
ball straight down and shoot the other off the table, the y motion would be identical (Pop and drop).
Any motion– any displacement or velocity or acceleration –can be broken down into its x and y
components. The important thing is that the x and y motions can be analyzed separately. What
you will notice, however, is that time is the variable linking the two sets of equations.
Ex. 9-1
A ball is shot from a toy gun with an
initial velocity of 6.7 m/s horizontally. If the gun is
1.16 m off the floor
a) how long does the ball take to hit the floor and
b) how far from the starting point does the ball
travel?
For our two times, let’s take the naughts to be the
point where the ball has just been fired from the gun
and the final time to be just before the ball hits the
floor.
+y
+x
Remember that we are working with constant acceleration. This means that axo = ax and ayo = ay.
We need to establish a coordinate system. Let’s take x = 0 and y = 0 to be at the point where the ball is
at to. Then xo = 0 and yo = 0. ax = 0 and ay = -9.8 m/s2.
Remember that the motions are totally independent of each other: the time it takes the ball to reach the
bottom is the same regardless of the horizontal motion. The t that we calculate will be the time it takes
the ball to reach the ground. So realize that we need to solve for time from the y-motion. (How long
does it take a ball to fall from this height? The x-motion doesn’t affect this! We will then move time
to the x-motion side of the problem and use it to solve for how far the ball has traveled in the xdirection.
Known: vyo = 0 m/s
Solve: t
ay = -9.8 m/s2
y = -1.16 m
1
y = v y 0t + a y t 2
2
1
y = ayt 2
2
2y
2(−1.16m)
t=
=t =
= 0.49 s
ay
−9.8 sm2
Not Involved: vy
Now that we have the time, realize that this time can be used in the x equations to figure out how far
the ball travels.
x = vx 0 t
x = vx 0 t
x = ( 6.7 ms ) ( 0.49s ) = 3.3 m
Ex. 9-2 A ball thrown from a building horizontally with a velocity of
4.2 m/s strikes the ground 16.0 m from the building. How tall is the
building?
+y
+x
We will again need to use the fact that the vertical and horizontal
components of motion are entirely independent. Let’s place the origin
of our coordinate system right where the ball leaves the edge of the
building. We can use the x-motion to solve for the time in the air and
then use this time in the y-motion and ask “from what height does it
take an object this long to fall to the ground?”.
x-motion
Since ax=0, we can simply use x = vx t
y-motion
Known: vyo = 0 m/s
ay = -9.8 m/s2
t = 3.8 s
Solve: y
t=
x 16.0 m
=
= 3.8 s
vx 4.2 m
s
Not Involved: vy
1
y = v0 y t + a y t 2
2
m⎞
1
1⎛
2
y = a y t 2 = ⎜ −9.8 2 ⎟ ( 3.8s ) = −71 m
s ⎠
2
2⎝
You Try
It!
An archer shoots an arrow horizontally at a target 15 m away. The arrow is aimed
directly at the center of the target, but it hits 52 cm lower. What was the initial speed of
the arrow?
Ex. 9-3 A cougar leaps horizontally from the top of a cliff with an initial velocity of 8.25 m/s. The
cliff is 6.43 m tall. What are location and velocity when the cougar impacts the ground?
We know the path is parabolic. We can calculate the time it takes for an object to fall from this
height.
2 ( −6.43m )
2y
=
= 1.15 s and from this the horizontal distance. x = v x t = ( 8.25m / s ) (1.15 s ) = 9.5m
m
a
−9.8 2
s
To calculate the velocity at the impact point we will need to know the x and y components of velocity
at that point. The x velocity is a constant 8.25 m/s in this problem since ax = 0. The y velocity it the
speed an object has after falling 6.43 m.
v y2 = v02 y + 2a y y
t=
m⎞
m
⎛
v y = 2a y y = 2 ⎜ −9.81 2 ⎟ ( −6.43 m ) = −11.2
s ⎠
s
⎝
Thus we need to vectorially combine vx=8.25 m/s and vy=-11.2 m/s. Using the Pythagorean Theorem:
2
2
m⎞ ⎛
m⎞
m
⎛
v = vx2 + v y2 = ⎜ 8.25 ⎟ + ⎜ −11.2 ⎟ = 13.9
s⎠ ⎝
s⎠
s
⎝
m
8.25
opp
s = 36.3o
θ = tan −1
= tan −1
m
adj
11.2
s
vx = 8.25 m/s
vy = 11.2 m/s
v = 13.9 m/s
θ
You Try
It!
A crow is flying horizontally with a constant
speed of 2.7 m/s when it releases a clam from
its beak. The clam lands on the rocky beach
2.10 s later. Just before the clam lands what is (a) the horizontal component of velocity and (b) its
vertical component of velocity? (c) How would your answers to parts (a) and (b) change if the speed of
the crow were increased?
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.
x 1.50 m
m
=
= 1.20
= v0 x
t 1.25 s
s
1.20 ms
v
v0 = 0 x =
= 1.57 m s
cos θ cos 40.0°
vx =
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 =
and at that time vy = 0.
1
(4.50 s) = 2.25 s,
2
So considering the motion from that time on and since vy = v0 y − gt,
FG
H
IJ
K
m
m
v0 y = 9.81 2 (2.25 s) = 22.07
s
s
m
22.07 s
So, v0 =
= 24.8 m s .
sin 63.0°
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?
voy = v0 sin θ
m⎞
⎛
= ⎜ 2.25 ⎟ sin ( 35° )
s⎠
⎝
m
= 1.29
s
Known: v0y = 1.29 m/s
Solve: y
NI: vy
2
a = -9.81 m/s
t = 1.60 s
1
y = voy t + at 2
2
m⎞
m⎞
1⎛
2
⎛
= ⎜ 1.29 ⎟ (1.60s ) + ⎜ −9.81 2 ⎟ (1.60 s )
s⎠
s ⎠
2⎝
⎝
= −10.5 m
You Try
It!
What was the girl’s greatest height above the water?
v y2 = v02 y + 2ay
2
m⎞
⎛
− ⎜1.29 ⎟
2
−v
s⎠
y = 0y = ⎝
= 0.08 m
m⎞
2a
⎛
2 ⎜ −9.81 2 ⎟
s ⎠
⎝
So adding this to 10.5 m gives only 10.6 m.
Ex. 9-4 : A football is kicked at an angle of 30 degrees to the horizontal with an initial velocity of 24.0
m/s. How far does it go?
At t
At to
x0 =0
y0 =0
x =R
y=h
vx0 = vo cos θ
vy0 = vo sin θ
vx =?
vy= ?
ax = 0
ay = -9.8 m/s2
vx0 = vo cos θ = (24.0 m/s cos(30)) = 20.8 m/s
vy0 = vo sin θ = (24.0 m/s sin(30)) = 12.0 m/s
At the top of the motion, vy = 0 (note that vx is not 0)
v y = v y0 + a y t
v y − v y0
ay
=t=
0 − 12.0 ms
− 9.8 sm2
= 1.22s
From this, we can find out how far it goes in the x direction:
1
x = x 0 + v x0 t + a x t 2
2
x = v x 0 t = 20.8 ms (1.22s ) = 25.4m
This is half as far as it goes, so the total length is 50.8 m
Note that the time in the air T = 2.44 s is T =
2v0 sin θ
g
b g
And the total distance traveled R (the horizontal range) is R = v0 cosθ T
Substituting the first equation into the second yields the horizontal range formula:
R=
2v02
v2
sin θ cosθ = 0 sin 2θ
g
g
Which in this example yields
b
c
g
2
24m / s
v2
R = 0 sin 2θ =
sin 60o = 50.9m
g
9.8m / s 2
h
Now we can ask the question: At what angle should one kick a ball to get the maximum horizontal range? We want the sin
function to have its maximum value of 1 which occurs when the argument is 90°. If 2θ = 90°, then θ = 45°.
The maximum height can also be defined.
v y2 = v02 y + 2ay
ymax =
v02y
2a
=
v 02 sin 2 θ
2g