Projectile Motion
when an object that moves through space is
acted upon by Earth's gravity
Ex. A football player kicks a football through the end
zone for a field goal
Of course there is an initial velocity, but
then only gravity....
Now we have 2 dimensions and we
break the initial velocity into its 2
components, horizontal and vertical
IMPORTANT ;
g only acts vertically, so the
vertical velocity will change
but the
horizontal velocity remains constant.
if the initial velocity is 20 m/s
at an angle of 15 degrees what are
the horizontal and vertical
components?
horizontal componenet
20 cos 15 = 19 m/s
vertical component
20 sin 15 = 5.2 m/s
Once we have the components:
we can use these values in the same equations we used for
rectilinear motion to find variables such as time, distance and
velocity.
Keep in mind the final velocity may also have 2 components and
you may need to find the resultant vector.
Let's start with an example that is launched horizontally.
That means, there is no initial velocity for the y component, only
the horizontal component.
A marble rolls of the table with a horizontal velocity of 1.2m/s
It lands in a cup 0.51m from the table's edge.
How high is the table?
Horizontal
vi = 1.2 m/s
vf = 1.2 m/s
di = 0m
df = 0.51m
knowing the velocity is constant I know that v = d/t so I can
find t
t = d/v = 0.51m/ (1.2m/s) = 0.43s
Now I have time
Vertical
vi = 0 m/s
g = -9.8 m/s2
di = x
df = 0m
t = 0.43s
What forumal?
df = di + vit + 1/2at2
0m = x + 0*t -4.9t2
x = 4.9*(.43)2
x = 0.91m
The table is 0.91 m high
Helpful Hints:
It is common to put the origin of the
cartesian plane at ground level where x =
0 is at the launch site
Projectile objects leaving the ground will
reach the peak height at half the total
time
If taking off from a different level
(height), the time it takes to return to it's
original height is twice the time it takes to
reach the peak height
Emmanuel Zacchini, the famous human cannonball, was fired out of a
cannon with a speed of 24.0 m/s at an angle of 40.0 degrees to the
horizontal. If he landed in a net 56.6 m away at the same height from
which he was fired, how long was Zacchini in the air?
Find the horizontal speed, and given the distance find t.
Vh = 24.0 m/s cos 40 = 18.4 m/s
t = d/v (velocity is constant) =
56.6m/ 18.4 m/s =3.08s
Evel Knievel, successfully jumped 69.5 m over a Grand Canyon
gorge. Assuming that he started and landed at the same level
and was airborne for 3.66s, what height from his starting point
did this daredevil achieve?
Knowing the max height was at the halfway point, we know that
t = 1.83s
g= -9.8m/s2
What equation
df = di + vit + 1/2at2
If we say vi = 0 at the halfway point
then di= x and df = 0
Then 0 = x + 1/2(-9.8m/s2)(1.83s)2
x = 16.4m
so the motorcycle reached a height of 16.4m
Ferdinand the frog is hopping from lily pad to lily pad in search of a
good fly for lunch.
If the lily pads are spaced 2.4 m apart, and Ferdinand jumps with a
speed of 5.0m/s, taking 0.60 s to go from lily pad to lily pad, at what
angle must Ferdinand make each of his jumps?
df= 2.4
di = 0m
vi = 5.0m/s
t = 0.60s
horizontal
v =d/t 2.4/0.69 = 4 m/s
The horizontal component of vi = vicosθ = vx
5.0 m/s * cos θ = 4 m/s
cos θ = 4/5
-1
θ = cos (4/5) = 37 degrees to horizontal
We often find the time first and then plug the time in to find other
variables.
Let's look at the examples in the textbook p 252-253
A soccer player kicks the ball with a velocity of 20.0 m/s at a 25.0
degree angle above the horizontal
How long does it take for the ball to reach its maximum
height?
A soccer player kicks the ball with a velocity of 20.0 m/s at a 25.0
degree angle above the horizontal
How long does it take for the ball to reach its maximum
height?
vix = 20.0 cos 25.0 v is constant
viy = 20.0 sin 25.0 v2 = 0m/x
What formula?
#1 Vf = Vi + at
0m/s = 8.45m/s - 9.8t
-8.45/-9.8 = t
t = 0.862s
What is the balls flight time?
We know it should be 2 * 0.86 seconds or 1.72 seconds
We can also use the formula
yf = yi + vit +1/2at2
Yf will be 0 again (it is back on the ground)
0 = 0 + 8.45m/s t - 4.9 m/s2t2
0 = t(8.45 -4.9t)
t = 0 (when the ball was kicked
9.45/4.9 = t
t= 1.72s
What distance does the ball travel horizontally?
We know the time is 1.72s
The horizontal velocity is 20.0m/s cos 25.0
= 18.1m/s (and its constant so vi and vf = 18.1m/s)
v = d/t and d = vt so d = 18.1m/s*1/72s = 31.2m
How do we get the resultand vector and the angle?
­8.45m/s
θ
18.1m/s
Pythagorean theorem:
find the hypotenuse = 20.0m/s
tan θ = -8.45/181
θ = -250
3600
-
250 = 3350 from the horizontal.
Example B:
Here is the question: Try it and check your answer in the text book.
From a roof 50.0m high, a ball is thrown with a velocity of 5.00m/s at 25.0o above the horizontal. List all your knowns for vertical and horizontal:
a) how long does it take for the ball to reach the ground?
( hint: you will need to use the quadratic equation: use the positive value of t)
b) what is the magnitude of the ball's velocity at the moment it touches the ground? What is the direction of the velocity vector at this moment
c) what is the maximum height reached by the ball?
Problems:
section 11.2, 251 all
Next time;
if we changed only the angle of the launch, what would happen to the maximum height?
and the horizontal distance?
design your own lab......
d) what is the value of the ball's velocity at the moment it touches the
ground?
give the magnitude and direction....
(yes it is a vector with an x component and a y component)
we know vxf = 18.1 m/s
vyf = ?
we have a and t and vi so we can use:
vf = vi + at
= 8.45m/s + (-9.8m/s2)(1.72s) = -8.45m/s (directed down)
­8.45m/s
18.1m/s
How do we get the resultand vector and the angle?