Transcript Projectile Motion - Conroe High School


Objects launched are projectiles

balls, bullets, arrows, space ships…
The PATH a projectile follows is the
TRAJECTORY
 The trajectory for a projectile is a PARABOLA

We will
ignore wind
resistance
Motion for a Projectile is
described in terms of…
Position
 Velocity
 Acceleration
All three of these are vector quantities and will be
represented with arrows like the previous topic
We must remember that horizontal and vertical
velocities (components) of a projectile are
INDEPENDENT!



Drop a ball
Accelerates
Downward
Roll a Ball
Rolls across
Constant Velocity (No friction)

Something thrown through the
air will have both a down
(vertical) and across (horizontal)
velocity
1.
2.
accelerate down (due to gravity)
go across at constant velocity(we
assume no air resistance)
 Each velocity
will NOT affect the
other (Independence)
Vh (Vx)
Vx is Constant
Vv (Vy)
Vy is changing
(accelerating b/c of
gravity)
What is the time of flight?
What is the Vx?
Equations
Past
D = Vt
D= Vit + ½ at2
Vf= Vi + at
Present
X = V xt
Y = Vyi t+ ½ gt2
Y = ½ gt2
t2 = 2y/g
Vyf = Vyi + gt
Sample Problem

A stone is thrown horizontally at +15
m/s from the top of a cliff 44m high.
1.
2.
3.
How long does the stone take to reach the
bottom of the cliff?
How far from the base of the cliff does the
stone strike the ground
Sketch the trajectory of the stone
15 m/s
Given
Vx = 15m/s
44m
Vyi = 0 m/s
G= -9.8m/s2
Formula
X = Vxt
Y = Vyt + ½ gt2
y = -44m
SOLUTION
t= 3sec
44m
45m
a. t2 = 2 y/g
t2 = 2(-44)/ -9.8
t = √9 t= 3sec
b. X= Vxt
= (15 m/s) (3s)
= 45m
Practice Problems
1.
2.
A stone is thrown horizontally at a speed of +5.0m/s
from the top of a cliff 78.4m high.
a. How long does it take the stone to reach the
bottom of the cliff?
b. How far from the base of the cliff does the
stone strike the ground?
A steel ball rolls with a constant velocity across a
tabletop .950m high. It rolls off and hits the ground
+.352m horizontally from the edge of the table.
How fast was the ball rolling?
Homework P. 102
1-3
Page 118 K and L
Objects Launched at Angles
When we analyze a
trajectory, we should see
what happens to vertical and
horizontal components to
understand the motion
Page 118 J
Find component velocities on the way
up.
 Find Resultant velocities on the way
down.

Shallow Angle
Steep Angle
Use Vector Knowledge to
Solve for
Vx & Vy
 The launching angle and the velocity
determine how far the object will travel.
 When the launching speed or force is the
same, the angle alone will determine the
range.
 RANGE: how far (horizontally) a projectile
goes
NOTE
For working with angled trajectories, the
launching height is usually the same as the
landing height When it is DY = 0!

Finding Time for angled
launches

Time =
2Vy
g
Only works when landing/launching
heights are the same. I call this
the “hangtime” formula.
Example Problem for a projectile shot at an angle…

A small metal ball is shot with a velocity
of 4.47m/s at an angle of 66° above the
horizontal.
a. How long does it take the ball to
land?
b. How high did the ball fly?
c. What was the range?
Formula
Given
y = Vyt + ½ gt2 A. when land y=0
2
0
=
V
t
+
½
gt
y
X=V t
Vi = 4.47m/s
0 = 66°
4.47
66
Solution
x
Vy
Vx
-Vyt = ½ gt2
t = -2Vy/ g
Vy = 4.47(sin 66) = 4.08m/s
t = -2(4.08)/-9.8
Vx = 4.47(cos 66) = 1.82 m/s
= .83 sec
B. max height =½t
y = Vyt + ½ gt2
y = (4.08)(.417) + ½ (-9.8)(.417)2
y = 1.7 – 0.852
C. Flight time = .833sec
y = 0.85m
X = Vxt
= 1.82m/s (.833sec)
X = 1.52m
Practice problem 5




A softball is thrown with an initial velocity of
27m/s at an angle of 30o from the horizontal.
A. Find the total time the ball is in the air.
B. find the horizontal displacement of the
ball.
C. Find the maximum height for the ball.
27 m/s
30°
Vy
Vx
Given
Vx = 23.38
Vy = 13.5
g = -9.8m/s
½ t= 1.38
A. t = -2Vy/ g
B. X = Vxt
= -2(13.5)/- 9.8
= 23.38(2.76)
= 2.76s
=65.5m
C. y = Vyt + ½ gt2
= 13.5(1.38) + ½ (-9.8)(1.38)2
18.63 + (-9.33)
= 9.3m
Problem 6
All the given information of this problem
is the same as #5. The only difference
is the launch angle is now 60o.
 A. find time
 B. Find range (X)
 C. Find max height (Y).

27 m/s
60°
Vy
Vx
Given
A. t = -2Vy/ g
B. X = Vxt
Vx = 13.5
-2(23.38)/-g
13.5(4.78)
Vy = 23.38
= 4.78s
=64.5m
g = -9.8
½ t = 2.39s
C. Y = Vyt+ ½ gt2
= 23.38 (2.39) + ½ (-9.8) 2.392
=27.9m
The Range Formula
 A fast way to find range when initial
velocity and angle are given (angle
should be with the horizontal)
Vi2 sin(2 0 )
Can also
Find Vi
R=
2
Vi
=
g
g R
sin (2 0 )
Example

Find the range for an object shot at 50° with
an initial velocity of 5m/s
Given
Vi = 5m/s
0 = 50°
R=?
g = 9.8
Formula
R = Vi2 Sin 2 0
g
Solution
= (5)2 sin (2x50°) / 9.8
+ 24.62/ 9.8 =
2.51m
p.104 3E #’s 5,2,1
p.114-115
30, 32, 34, 36, 39
Complimentary Angles = same ranges
450 gives the longest range for a projectile shot
without air resistance. While the ranges may be the
same, the time in the air won’t.
What are satellites?
Natural
 Artificial

How do satellites stay in orbit even
though they have no propulsion?

Isaac Newton saw the answer to this by
thinking about the Moon. He also knew
over 400 years ago that it would be
possible to orbit the Earth if mankind
could go fast enough. He also
predicted mankind would travel to the
Moon.
For low Earth orbit(200 to 500miles high)

Tangential Velocity of 8000m/s

Vt will change with altitude.
(17,500mph)
The object (satellite) will stay in orbit.
 Every second it falls down 4.9m and
moves downrange (tangentially)
8000m. The resultant path mirrors the
curvature of the Earth.

Vectors and Projectiles Review
Mr. Crabtree is standing on a perfectly good bridge
in Zimbabwe when he decides to jump off. He is
tied by his ankles to a bungee cord for safety. For
5.2 seconds, your teacher is in free fall. He
jumped horizontally at 4 m/s as he left the bridge.
The bridge is 150m above the Zambezi River.
What was his vertical displacement at the end of
the free fall? [neglect wind resistance]
Find horizontal displacement from the
base of the building.
Find the time the ball is in the air
10m
20m