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PROJECTILE MOTION 06/06/2022 IB Physics (IC NL) 1
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### Transcript of Projectile motion

04/08/2023 IB Physics (IC NL) 1

PROJECTILE MOTION

Projectile Motion

Motion in Two Dimension

04/08/2023 IB Physics (IC NL) 2

04/08/2023 IB Physics (IC NL) 3

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.

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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

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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 projectile2. θ = 90 vertical projectile (studied earlier)3. θ = θ which is the general case.

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HORIZONTAL PROJECTILES•Horizontal Projectiles are easiest to work with•only formula used in horizontal (x) direction is:

Remark: in case of horizontal projectiles select the direction of the y-axis to be downward.

x = ut

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HORIZONTAL PROJECTILES

•Horizontal Projectiles are the most basic•only formula used in horizontal (x) direction is:

constant speed!

x = ut

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HORIZONTAL PROJECTILES•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 = gt2

t1

t2

t3

t4

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start by drawing a picture:

EXAMPLEA 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 givens1890 m.s

1.7 cm

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EXAMPLE

What is the horizontal distance between the rifle and the bull’s-eye?

want:

1890 m.s

1.7 cm

X Y

xu y

ay

yu

1890 m.s 1.7 cm

2

m9.8

s10 m.s

dx horizontal distance

1m0.017m

100cm

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which equation do we use?

EXAMPLE

2y y

1y a t u t

2 use to find time

rewrite equation for t

y

2yt

a

0

2(0.017)

9.8

0.059 s

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Use t and ux to solve for x

EXAMPLE

xx u t (890)(0.059)

52.4 m

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Click icon to add online image

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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

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NON-HORIZONTAL PROJECTILES

•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

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NON-HORIZONTAL PROJECTILES

•need to calculate initial ux and uy

v

xu

yu

u cos

u sin

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VISUALIZING PROJECTILES•first enter vectors•focus on ux

vx = ux is constant the whole flight!

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VISUALIZING PROJECTILES•first enter vectors•focus on vx•focus on vy

vy decreases as it

rises!by how much per second?

no vy at the top!

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VISUALIZING PROJECTILES

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

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Mathematical analysis

Applying Newton’s 2nd law: ∑=mThe only force acting in the absence of air resistance is mThen m=m and = Resolving along x-axis: Resolving along y-axis:ax =0 ay = -g

vx=ucosθ vy = -gt + usinθ

x =ucosθt y = -gt2 +usinθtThus motion along x is thus motion along y is URM UARM

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Importance of time

Equation of trajectory:Eliminate t in x and substitute for t in y to get the following:

y = - 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 =

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 = Remember: max x when θ = 45o

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LET’S ANALYZE THE JUMP

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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.

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Click icon to add online image

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.

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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 thepath is not a true parabola.

22

In the case of air resistance, the path of a high-speedprojectile falls below the idealized path and followsthe solid curve.

Computer-generated trajectories of a baseball with and without drag.

23