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8/20/2019 Lecture 1 - Projectiles
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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
8/20/2019 Lecture 1 - 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$%
5 10 15 20 25 30 35 40
5
10
15
20
x (m) horizontal distance
y (m) height
#" cos " ms$%
#" sin " ms$%
.- ms$# 1 ve
"
.-
÷−
#"cos"
#"sin"
÷
8/20/2019 Lecture 1 - Projectiles
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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
x (m) horizontal distance
y (m) height
#" cos " ms$%
#" sin " ms$%
.- ms$# 1 ve
%"
%.0# .-t
= ÷−v
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Horizontal Vertical sh = #" 3cos "4t sv = #" 3sin "4t + 5.t
#
sh = %"t sv = %.0# + 5.t #
&sing the suvat e2uation , ' may define the hori)ontal
and vertical positions, sh and sv. 3or x and y4
5 10 15 20 25 30 35 40
5
10
15
20
x (m) horizontal distance
y (m) height
#" cos " ms$%
#" 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
x (m) horizontal distance
y (m) height
vh = %"
H + ma7. height
vv = "
1 ve
%.0#%. s
.-t = =
8/20/2019 Lecture 1 - Projectiles
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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)
8/20/2019 Lecture 1 - Projectiles
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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
x (m) horizontal distance
y (m) height
v sin θ .-
v cos θ
v
θ
1 ve
8/20/2019 Lecture 1 - Projectiles
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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.=
8/20/2019 Lecture 1 - Projectiles
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8/20/2019 Lecture 1 - Projectiles
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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$#
1ve
8/20/2019 Lecture 1 - Projectiles
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a4 =ow long does the pebble take to reach the house>
vh %%.: cos " :.:
ah ", hence,
".: m
%." m
%.: m
%." m
%." m
%" m
%." m
"o
%%.: ms$%
downstairs window
CulietBs window.- ms$#
1vedistance %"
time %.5 sspeed :.:
= =
8/20/2019 Lecture 1 - Projectiles
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b4 ?oes the pebble hit CulietBs window>
av + .-, uv %%.: sin " ., t %.5, so %, vv >
CulietBs window is E 0 m high and
F 5 m high, so the stone hits her window!
".:m
%."
m
%.:
m
%."
m
%."
m
%" m
%." m
"o
%%.: ms$
%
downstairs
window
CulietBs
window.- ms$#
1ve
#
"
#
%use
#
% . %.5 ".: .- %.5
0.:"m
v
v
v
s s ut at
s
s
= + +
= + × + × − ×
=
8/20/2019 Lecture 1 - Projectiles
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c4
use v u 1 at
vv . 1 +.- 9%.5 +."#
?raw a vector diagram!
8peed is the
magnitude of v
8peed of pebble
when it hits
the house is, .%0
".:
m
%."
m
%.:
m
%." m
%."
m
%"
m
%." m
"o
%%.: ms$
%
downstairs
window
CulietBs
window.- ms$#
1v
e
."#
:."
v# #:.: ."# .%0= + =v
8/20/2019 Lecture 1 - Projectiles
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8/20/2019 Lecture 1 - Projectiles
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A ball is kicked from ground level over hori)ontal ground. 't leaves at aspeed of #: ms$% 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
8/20/2019 Lecture 1 - Projectiles
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A ball is kicked from ground level over hori)ontal ground. 't leaves at a
speed of #: ms$% at an angle of θ to the hori)ontal such that cos θ ".
and sin θ ".#-.i4 ?erive e7pressions for the hori)ontal and vertical displacement at
time t .
Jositive upwards.
@ertical* uv = #" sin θ #: 9 ".#- , av = – .-
&se*
=ori)ontal* uh = #" cos θ #: 9 ". #5, ah "
Again use*
# #% 5.#
s ut at t t = + ⇒ = −
#% #5#
s ut at & t = + ⇒ =
8/20/2019 Lecture 1 - Projectiles
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A ball is kicked from ground level over hori)ontal ground. 't leaves at aspeed of #: ms$% 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 = =
8/20/2019 Lecture 1 - Projectiles
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8/20/2019 Lecture 1 - Projectiles
22/25
A ball is kicked from ground level over hori)ontal ground. 't leaves at a
speed of #: ms$% at an angle of θ to the hori)ontal such that cos θ ".
and sin θ ".#-.iv4 6ind, when t %.#:* A4 the vertical component of velocity of the
ball. I4
8/20/2019 Lecture 1 - Projectiles
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A ball is kicked from ground level over hori)ontal ground. 't leaves at a
speed of #: ms$% at an angle of θ to the hori)ontal such that cos θ ".
and sin θ ".#-.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
& & & & & &
& & & &
= − ==
= − = − = − ÷ ÷ ÷
= × − = −
8/20/2019 Lecture 1 - Projectiles
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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 ".#-.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
& &
& &
= −
= =
=
8/20/2019 Lecture 1 - Projectiles
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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