Projectile motion and its uses in daily life and its basic use.

PROJECTILE MOTION
8/20/2017 IB Physics (IC NL) 1

Projectile Motion
Motion in Two Dimension

Topic objectives
 State the independence of the vertical and the horizontal
components of velocity for a projectile in a uniform field.
 Describe and sketch the trajectory of projectile motion as
parabolic in the absence of air resistance.
 Describe qualitatively the effect of air resistance on the
trajectory of a projectile.
 Solve problems on projectile motion.
 Remarks:
1. Proof of the parabolic nature of the trajectory is not required.
2. Problems may involve projectiles launched horizontally or at any angle above or below the
horizontal. Applying conservation of energy may provide a simpler solution to some problems
than using projectile motion kinematics equations.

What is a projectile?
 When a body is in free motion, (moving
through the air without any forces apart
from gravity and air resistance), it is called a
projectile
 Normally air resistance is ignored so the only
force acting on the object is the force due to
gravity
 This is a uniform force acting downwards

Types of projectiles
There are three types of projectile depending on
the value of the angle between the initial velocity
and the x-axis.
1. θ = 0 horizontal projectile
2. θ = 90 vertical projectile (studied earlier)
3. θ = θ which is the general case.

8/20/2017 IB Physics (IC NL)
HORIZONTAL PROJECTILES
•Horizontal Projectiles are easiest to work
with
•only formula used in horizontal (x)
direction is:
6
Remark: in case of horizontal
projectiles select the direction
of the y-axis to be downward.
x = ut

•Horizontal Projectiles are the most basic
•only formula used in horizontal (x)
direction is:
constant speed!
7
x = ut

•vertical (y) direction is just freefall
•all of the initial velocity is in the x
direction
•So, u
• uy = 0
•since uy is in
freefall,
• a = +9.8 m.s-2
• y =
1
2
gt2

t1

t2
t3
t4
8

start by drawing a picture:
EXAMPLE
A person decides to fire a rifle horizontally at a
bull’s-eye. The speed of the bullet as it leaves the
barrel of the gun is 890 m.s-1. He’s new to the
ideas of projectile motion so doesn’t aim high and
the bullet strikes the target 1.7 cm below the
center of the bull’s-eye.
What is the horizontal distance between the rifle
and the bull’s-eye?
label the explicit givens
1
890 m.s

1.7 cm
9

EXAMPLE
What is the horizontal distance between the rifle
and the bull’s-eye?
want:
1
890 m.s

1.7 cm
X Y
xu  y 

ay 
yu 
1
890 m.s
1.7 cm
2
m
9.8
s 1
0 m.s
dx
horizontal distance
1m
0.017m
100cm
 
 
 
10

which equation do we use?
EXAMPLE
2
y y
1
y a t u t
2
 use to find time
rewrite equation for t
y
2y
t
a


0
2(0.017)
9.8


 0.059 s
11

Use t and ux to solve for x
EXAMPLE
xx u t (890)(0.059)

 52.4 m
12

NON-HORIZONTAL PROJECTILES
• vx = ux is still constant
• uy is also constant
•only difference with non-horizontal is
that vy is a function of time
u
14

•Angled Projectiles require a little work
to get useful u
•u has an x and y component
•need to calculate initial ux and uy
u
15

•need to calculate initial ux and uy

v
xu
yu
 u cos 
usin 
16

VISUALIZING PROJECTILES
•first enter vectors
•focus on ux
vx = ux is
constant the
whole flight!
17

•first enter vectors
•focus on vx
•focus on vy
vy decreases as it
rises!
by how much per second?
no vy at the top!
18

19

Boundary conditions: at t = 0, x0 = y0 = 0
ux = ucosθ and uy = usinθ

Mathematical analysis
Applying Newton’s 2nd law: ∑ 𝐹=m 𝑎
The only force acting in the absence of air resistance is m 𝑔
Then m 𝑔=m 𝑎 and 𝑔 = 𝑎
Resolving along x-axis: Resolving along y-axis:
ax =0 ay = -g
vx=ucosθ vy = -gt + usinθ
x =ucosθt y = -
1
2
gt2 +usinθt
Thus motion along x is thus motion along y is
URM UARM

Importance of time
 Equation of trajectory:
Eliminate t in x and substitute
for t in y to get the following:
y = -
𝑔
2𝑢2 𝑐𝑜𝑠2θ
x2 + tanθ x
This is an equation of parabola
 For maximum height reached,
Remember vy at top is zero. Get
t from vy=0 then plug t in y you
get ymax =
𝑣2 𝑠𝑖𝑛2θ
2𝑔
Whatever you need to calculate look for time which is common for x and y
The range is the xmax at
horizontal level of x-axis. To
calculate xR or xmax, take y
= 0 since displacement, get
t from y =0 you get two
times, one is zero (trivial,
lunching point) and
substitute for the other t in
x to get:
xR = xmax =
𝑣2 𝑠𝑖𝑛2θ
𝑔
Remember: max x when θ = 45o

LET’S ANALYZE THE JUMP
23

VARIED ANGLES
•which projectile angle shoots highest?
•larger θ means faster uy
•which projectile angle shoots farthest?
•45° has perfect balance of fast vx and long flight
time.
24

If projectiles are launched at the same speed, but at different
angles, the height and range is of the projectile are affected.

Solving Problems Involving Projectile
Motion
1. Read the problem carefully, and choose the
object(s) you are going to analyze.
2. Draw a diagram.
3. Choose an origin and a coordinate system.
4. Decide on the time interval; this is the same in
both directions, and includes only the time the
object is moving with constant acceleration g.
5.Examine the x and y motions separately.

6. List known and unknown quantities.
Remember that vx never changes, and that
vy = 0 at the highest point.
7. Plan how you will proceed. Use the
appropriate equations; you may have to
combine some of them.
Solving Problems Involving Projectile
Motion (cont.)

When the effect of air resistance is significant,
the range of a projectile is diminished and the
path is not a true parabola.
22

In the case of air resistance, the path of a high-speed
projectile falls below the idealized path and follows
the solid curve.
Computer-generated trajectories of a baseball with and
without drag.
23

Projectile motion and its uses in daily life and its basic use.

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