Lecture 1 - Projectiles

8/20/2019 Lecture 1 - Projectiles

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The picture below shows the path through the air of a jet of water.This path is known as a Trajectory.

This path would be similar if you propelled a ball by kicking/hitting

it. i.e. when it has been hit or kicked, the only force acting is gravity!

You can see that the Trajectory is a Parabola.

Any object moving

through the air in

this manner, solely

under the influence

of gravity is

called a Projectile!

Projectiles


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The projectile is a particle

The projectile is not under power

The air does not affect the motion

The only force acting is gravity, vertically downwards onsider a golf ball being chipped from ground level at a velocity of

The diagram

would look like

this.

Projectiles – Modelling Assumptions

5 10 15 20 25 30 35 40

5

10

15

20

x (m) horizontal distance

y (m) height

"o

#" ms$%

% o#"ms "− ∠


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&sing what we learned previously, ' have now resolved the velocity

into two components. (ne hori)ontal, the other vertical.

This gives*

Horizontal Vertical Vector

Initial posn. " "

Accln. " + .- ms$#

Initial vel. #" cos " ms$% %" #" sin " ms$% %.0#

#" sin " ms$%

"o #" cos " ms$%

#" ms$%

5 10 15 20 25 30 35 40

5

10

15

20


y (m) height

#" cos " ms$%

#" sin " ms$%

.- ms$# 1 ve

"

.-

÷−

#"cos"

#"sin"

÷


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Horizontal Vertical

vh = #" cos " vv = #" sin " + .-t

vh = %" vv = %.0# + .-t

&sing the suvat e2uation. v u 1 at , ' may define the hori)ontal and

vertical velocities, vh and vv.

5 10 15 20 25 30 35 40

5

10

15

20


y (m) height

#" cos " ms$%

#" sin " ms$%

.- ms$# 1 ve

%"

%.0# .-t

= ÷−v


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Horizontal Vertical sh = #" 3cos "4t sv = #" 3sin "4t + 5.t

#

sh = %"t sv = %.0# + 5.t #

&sing the suvat e2uation , ' may define the hori)ontal

and vertical positions, sh and sv. 3or x and y4

5 10 15 20 25 30 35 40

5

10

15

20


y (m) height

#" cos " ms$%

#" sin " ms$%

.- ms$# 1 ve

#

%"

%.0# 5.

t

t t

= ÷−r

#%

# s ut at = +


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The Maximum Height ( H )

vh = %" 31)

vv = %.0# + .-t (2)

sh = %"t ( 3)

sv = %.0#t + 5.t # (4)

6rom (2)* vv = %.0# + .-t and at the ma7imum height e2uals )ero!

" %.0# + .-t

8ubstituting for t in (4) sv = %.0#t + 5.t #

sv = %.0#9%. + 5. 9%.# %:.0 m

;a7 height, H %:.0 m

5 10 15 20 25 30 35 40

5

10

15

20


y (m) height

vh = %"

H + ma7. height

vv = "

1 ve

%.0#%. s

.-t = =


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The Time of Flight (T ) and ange ( R)

vh = %" 31)

vv = %.0# + .-t (2)

sh = %"t ( 3)

sv = %.0#t + 5.t # (4)


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The Time of Flight (T ) and ange ( R) ( Alternative method for T)

vh = %" 31)

vv = %.0# + .-t (2)

sh = %"t ( 3)

sv = %.0#t + 5.t # (4)


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Horizontal!

uh v cosθvh = v cosθah " sh = v 3cosθ 4t

Vertical!

uv v sinθ vv = v sinθ + .-t sv = v 3sinθ 4t + 5.t #

General Projectile Euations

5 10 15 20 25 30 35 40

5

10

15

20


y (m) height

v sin θ .-

v cos θ

v

θ

1 ve


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A ball is projected hori)ontally at : ms$% from a window 5 m above the

ground. a4 how long does it take to hit the ground> b4 how far from

the window does it land> c4 what is the velocity, as a magnitude and

direction at the instant it hits the ground>

?raw a diagram

a4 @ertical* uv ", sv 5, a .-, t >

&se* s ut 1 !."at # ⇒ 5 5.t #

t =

b4 =ori)ontal* sh R uh t = : 9 "."5

R 5.:# m

Projectile Pro$lems

: ms$%

R m

5 m

.- ms$#

1 ve

5"."5s

5.=


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The diagram depicts omeoBs attempt to attract CulietDs attention by

throwing a pebble at her window. =e launches the pebble at a height of

% m above the ground, at a speed of %%.: ms$% , "o to the hori)ontal.

a4 =ow long does the pebble take to reach the house>

b4 ?oes the pebble hit CulietBs window>c4

E&am 'tle )uestion

".: m

%." m

%.: m

%." m

%." m

%" m

%." m

"o

%%.: ms$%

downstairs window

CulietBs window.- ms$#

1ve


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a4 =ow long does the pebble take to reach the house>

vh %%.: cos " :.:

ah ", hence,

".: m

%." m

%.: m

%." m

%." m

%" m

%." m

"o

%%.: ms$%

downstairs window

CulietBs window.- ms$#

1vedistance %"

time %.5 sspeed :.:

= =


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b4 ?oes the pebble hit CulietBs window>

av + .-, uv %%.: sin " ., t %.5, so %, vv >

CulietBs window is E 0 m high and

F 5 m high, so the stone hits her window!

".:m

%."

m

%.:

m

%."

m

%."

m

%" m

%." m

"o

%%.: ms$

%

downstairs

window

CulietBs

window.- ms$#

1ve

#

"

#

%use

#

% . %.5 ".: .- %.5

0.:"m

v

v

v

s s ut at

s

s

= + +

= + × + × − ×

=


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c4

use v u 1 at

vv . 1 +.- 9%.5 +."#

?raw a vector diagram!

8peed is the

magnitude of v

8peed of pebble

when it hits

the house is, .%0

".:

m

%."

m

%.:

m

%." m

%."

m

%"

m

%." m

"o

%%.: ms$

%

downstairs

window

CulietBs

window.- ms$#

1v

e

."#

:."

v# #:.: ."# .%0= + =v


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A ball is kicked from ground level over hori)ontal ground. 't leaves at aspeed of #: ms$% at an angle of θ to the hori)ontal such that cos θ ".

and sin θ ".#-.

i4 ?erive e7pressions for the hori)ontal and vertical displacement at

time t .

ii4 alculate the ma7imum height the ball reaches.

iii4 alculate the times when the ball is at half its ma7imum height and

the hori)ontal distance travelled between these two times.

iv4 6ind, when t %.#:* A4 the vertical component of velocity of the

ball. I4


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A ball is kicked from ground level over hori)ontal ground. 't leaves at a

speed of #: ms$% at an angle of θ to the hori)ontal such that cos θ ".

and sin θ ".#-.i4 ?erive e7pressions for the hori)ontal and vertical displacement at

time t .

Jositive upwards.

@ertical* uv = #" sin θ #: 9 ".#- , av = – .-

&se*

=ori)ontal* uh = #" cos θ #: 9 ". #5, ah "

Again use*

# #% 5.#

s ut at t t = + ⇒ = −

#% #5#

s ut at & t = + ⇒ =


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A ball is kicked from ground level over hori)ontal ground. 't leaves at aspeed of #: ms$% at an angle of θ to the hori)ontal such that cos θ ".

and sin θ ".#-.

ii. alculate the ma7imum height the ball reaches.Jositive upwards

v ", u , a +.-

&se* v# = u# 1 #as ⇒ " # 1 #9 + .- s 5 + %. s

E&am – 'tle )uestion

5#.: m

%.

s = =


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and sin θ ".#-.iv4 6ind, when t %.#:* A4 the vertical component of velocity of the

ball. I4


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and sin θ ".#-.v4 8how that the artesian e2uation of the trajectory is*

and find the range of the ball.".

3#5" 4:

& &= −

( ) ( )

#

# # #

# #

#

#

5. % #5 #

from # subs. for in ##5

5. %"

5. ".#5 #5 #5 #5 #5 #5

". ".%" #5 #5"

#5 :

t t E & t E &

E t t E

& & & & & &

& & & &

= − ==

= − = − = − ÷ ÷ ÷

= × − = −


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and sin θ ".#-.v. 6ind the range of the ball.

To find the range we needto find & when "

"." 3#5" 4

:

" or #5" $ "

#5"ange 05.0 m

& &

& &

= −

= =

=


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= @

'nitial

position " "

a " – g

u u & u cos % u u sin %

v v & u cos % v u sin % + gt

r & ut cos % ut sin % + L gt #

General Euations – velocit - positionx

g

%

u

(

1 ve

Lecture 1 - Projectiles

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Transcript of Lecture 1 - Projectiles