Lecture 07 - Projectile Motion

7/29/2019 Lecture 07 - Projectile Motion

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Chapter 3: 3

Projectile Motion



Exam I

Exam I (chap 1 -3) will be on Tuesday, Feb 5 during lecture

class.

Please bring your calculator, pencil, ruler & protractor.

Everything else (scratch paper, equations, constants, the exam,etc.) will be provided.

We will start right at 11:30 am. So, please be on time as extra

time cannot be given.



Projectile Motion

2D Motion of an object projected

into the air at an angle

Examples:

• A projectile hitting a target• A soccer ball being kicked

• A base ball being thrown

• A golfer hitting a golf ball

• An athlete long jumping

• An athlete diving into water



Three different

scenarios forprojectile motion

Object launched

from the ground at

certain launch angle

Object launched

from height at

certain launch angle

Object launched

from height at zero

launch angle



http://phet.colorado.e

du/sims/projectile-motion/projectile-

motion_en.html

http://phet.colorado.edu/sims/projectile-motion/projectile-motion_en.html












Things that we want to know about any

Projectile Motion

How High, H?

How Far, R?

How Long, tR?



We can answer these questions using

Equations of motion

The vertical and horizontal motions of an object are independent of

each other.

One can apply equation of motion to motion along x-direction and

y – direction separately .

The acceleration of the object along x –

axis is zero, a x = 0 m/s2

The acceleration of the object along y – axis is, a y = - 9.8 m/s2

But, first some facts about projectile motion:



We now apply equation of motion

motion along x-axis

a x = 0; xo = 0; vox = vocosq

v y = voy + a yt

y = yo + voyt + 1/2a yt 2

v y

2

= voy

2

+ 2a y(y –

yo)

vo –

initial launch velocityq – launch angle

vo

q

v x = vox + a xt

x = xo + voxt + 1/2a xt 2

v x2 = vox2 + 2a x(x –

xo)

a y = - 9.8 m/s2 ; yo = 0; voy = vosinq

motion along y-axis



motion along y-axis

a y = - 9.8 m/s2 ; voy = vosinq ; v y = 0

0 = vosinq + (-9.8)t

t = vosinq /9.8

t R = 2t = 2vosinq /9.8

v y = voy + a yt

How long, t R?

q



How far, R?

motion along x-axis

a x = 0 m/s2 ; xo = 0; x = R; vox = vocosq ; t = 2vosinq /9.8

R = 0 + (vocosq )(2vosinq /9.8)

x = xo

+ vox

t + 1/2a x

t 2

R = (vocosq )(2vosinq ) /9.8

R = vo2sin2q /9.8

q



How high, H? motion along y-axis

a y = - 9.8 m/s2 ; yo = 0; y = H ; voy = vosinq ; v y = 0

v y2 = voy

2 + 2a y(y –

yo)

0 = (vosinq )2 + 2(-9.8)(H – 0)

H = (vosinq )2 / (2 x 9.8)

q



A battle ship fires shells at two destroyers

simultaneously with same speed. Which shell hits first?

1 2 3 4

0% 0%0%0%

BA

1. A

2. B

3. They hit simultaneously

4. Need more information

t R = 2vosinq /9.8



The two balls start at the same height as shown. One is

dropped from rest while a second is fired horizontally

with a speed v. Which one hits the ground first?

1 2 3 4

25% 25%25%25%

1. Red Ball

2. Yellow Ball

3. Both hit at the same time

4. Depends on the mass of the ball



Projectile motion(scenario II)

motion along x-axis

a x = 0; xo = 0; vox = vocosq

motion along y-axis

a y = - 9.8 m/s2 ; yo = 0; voy = vosinq

vo – initial launch velocity

q – launch angle

Use equations of motion to determine:

1) How far?

2) How High?

3) How long?



Projectile motion (scenario III)

motion along x-axis

a x = 0; xo = 0; vox = vo

motion along y-axis

a y = - 9.8 m/s2 ; yo = 0; voy = 0

vo – initial launch velocity

q = 0 (launch angle)

Use equations of motion to determine:

1) How far?

2) How long?



Projectile motion: How the velocity vector

changes

Lecture 07 - Projectile Motion

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