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2 – Dimensional 2 – Dimensional KinematicsKinematics
PROJECTILE MOTIONPROJECTILE MOTION
(Right-click to pause presentation at any time)
PROJECTILE MOTIONPROJECTILE MOTION
The trajectory (path) of a projectile is parabolic.The trajectory (path) of a projectile is parabolic.
The vertical motion determines the time of flight.
PROJECTILE MOTIONPROJECTILE MOTION
V0V0YV0Y
V0XV0X
V1
V1X = V0X
V1Y = 0 m/s
V2V2
V2YV2Y
V2X = V0XV2X = V0X
The horizontal motion is uniform. (aX = 0 m/s2)
Vertical motion is accelerated. (aY= g = 9.80 m/s2)
g
SPECIAL CASE ASPECIAL CASE A
Projectile fired horizontally, θ = 0o
Vo
Δdy
Δdx (range)
Case ACase A
Horizontal motion (uniform)Horizontal motion (uniform)
d tx oxv ( cos )ov t d tx ov cos
NOTE : SINCE θ = 0o, cos θ = 1NOTE : SINCE θ = 0o, cos θ = 1
d tx ovWe need to use the vertical motion to find ΔtWe need to use the vertical motion to find Δt
Case ACase AVertical motion (to find Δt)
d t g ty oyv 12
2
Since θ = 0o and voy = vo sin θ, Since θ = 0o and voy = vo sin θ, sin θ = 0 and voy =0sin θ = 0 and voy =0
d g ty 12
2
td
gy
2(time of flight)(time of flight)
Case ACase A
Plug
td
gy
2
t
d
gy
2 into d tx ov d tx ov
d vd
gx oy
2
HOTLINK 1HOTLINK 1
SPECIAL CASE B
Projectile fired from “ground” level, Δdy = 0 m
Vo
Voy
g
Vox Δdx
Case BCase B
Horizontal motion (uniform)Horizontal motion (uniform)
d tx oxv ( cos )ov t d tx ov cos
We need to use the vertical motion to find ΔtWe need to use the vertical motion to find Δt
Case BCase BVertical motion (to find Δt)
d t g ty oyv 12
2
Since Δd = 0 mSince Δd = 0 m
0 12 t v g toy( )
0 012 v g ty
Case BCase B0 0
12 v g ty 0 012 v g ty
12 g t voy
tv
goy
2
(time of flight)(time of flight)tv
go
2 sin
Case BCase B
Plug Plug intointo d tx ov cos d tx ov costv
go
2 sintv
go
2 sin
d vv
gx oo
( )sincos
2
dv
gxo
2 2 sin cos
HOT LINK 2
Case C Case C The most general case.
Δdy
Vo
Vo
Δdy
Case CCase CVertical motion ( to find the time of flight)Vertical motion ( to find the time of flight)
d v t g ty oy 12
2 d v t g ty oy 12
2
Put into standard quadratic formatPut into standard quadratic format
0 12
2 g t v t doy y ( )
And apply the quadratic formulaAnd apply the quadratic formula
tv v g d
goy oy y
2 12
12
4
2
( )( )
( )
t
v v g d
goy oy y
2 12
12
4
2
( )( )
( )
Case CCase CSimplifying, we get :
tv v g d
goy oy y
2 2
t
v v g d
goy oy y
2 2
And since voy = vosinθAnd since voy = vosinθ
tv v g d
go o y
sin sin 2 2 2
t
v v g d
go o y
sin sin 2 2 2
(This is the equation for the time of flight)(This is the equation for the time of flight)
Case CCase C
Horizontal motion (to find the range)Horizontal motion (to find the range)
d v tx oxSince vox = vo cos θ ,Since vox = vo cos θ ,
d v tx o( cos )Now plug the expression for Δt into this equation.Now plug the expression for Δt into this equation.
Case CCase C
dv
gv v g dx
oo o y
cossin sin( ) 2 2 2
The use of “+” or “-” depends on the path of the projectile. The use of “+” or “-” depends on the path of the projectile. 1) If the projectile passes over the apex, use the “-” sign.1) If the projectile passes over the apex, use the “-” sign.
2) If it does not, use the “+” sign.2) If it does not, use the “+” sign.
{Since g is negative, use of the “-” sign gives a greatertime of flight.}{Since g is negative, use of the “-” sign gives a greatertime of flight.}
Case CCase C
For ease of calculation, this is usually written:For ease of calculation, this is usually written:
dv
gv v g dx
oo o y
cossin sin( ) 2 2 2
Use “+” signfor both
(not over apex)
Use “+” signfor both
(not over apex)
Use “-” sign for both(over the apex)
Use “-” sign for both(over the apex)
HOTLINK 3
Homework ProblemHomework ProblemA basketball player shoots a basket by launching the ball at an angle of 60.0o
from a location 1.0 m below the rim at12 m/s. How far is he from the basket
if he makes the shot? (hint : define up as +)
Practice the problems on theProjectile Motion Practice Sheets
The following pagescontain all of thepractice problems.
CHAPTER 7 PROJECTILE MOTION PROBLEMS
1. A ball falls from rest from a height of 490 m. a. How long does it remain in the air? b. If the ball has a horizontal velocity of 2.00 m/s when it begins its fall, what horizontal displacement will it have?
2. An archer stands 40.0 m from the target. If the arrow is shot horizontally with a velocity of 90.0 m/s, how far above the bull’s-eye must he aim to compensate for gravity pulling his arrow downward?
3. A bridge is 176.4 m above a river. If a lead-weighted fishing line is thrown from the bridge with a horizontal velocity of 22.0 m/s, how far has it moved horizontally when it hits the water?
4. A beach ball, moving with a speed of +1.27 m/s, rolls off a pier and hits the water 0.75 m from the end of the pier. How high above the water is the pier?
5. Carlos has a tendency to drop his bowling ball on his release. Instead of
having the ball on the floor at the completion of his swing, Carlos lets go with the ball 0.35 m above the floor. If he throws it horizontally with a velocity of 6.3 m/s, what distance does it travel before you hear a “thud”?
6. A discus is released at an angle of 45° and a velocity of 24.0 m/s. (assume Δdy = 0 m) a. How long does it stay in the air? b. What horizontal distance does it travel?
7. A shot put is released with a velocity of 12 m/s and stays in the air for 2.0 s. (assume Δdy = 0 m)
a. At what angle with the horizontal was it released? b. What horizontal distance did it travel?
8. A football is kicked at 45° and travels 82 m before hitting the ground. a. What was its initial velocity? b. How long was it in the air? c. How high did it go?
9. A golf ball is hit with a velocity of 24.5 m/s at 35.0° above the horizontal. Find :
a. the range of the ball. b. the maximum height of the ball.
10. A car moving at 120 Km/hr on a flat horizontal road loses control and careens off a cliff 75 m high into the valley below. Neglecting wind resistance, how far from the base of this sheer cliff does the car land?
11. You are a detective investigating an accident similar to the one described in problem #1 except that the cliff is 50 m high and the car impacts with the ground 20 m from the base. The speed limit was
50 Km/hr . How fast was the car going when it left the cliff? Could excess speed have contributed to the accident?
12. A baseball is struck with a bat at a height of 1.0 m giving it a speed of
60 m/s at an angle of trajectory of 30o. How far from home plate would the outfielder have to be in order to catch the ball at a height of 1.0 m? (Note: the distance to the outfield wall in this park is 110 m.)
13. A basketball player attempts to shoot a basket from mid-court (13 m from the basket) by making a jump shot and releasing the ball from a height of 3.0 m. If the basket is 3.0 m high and he gives the ball an angle of trajectory of 50o, at what speed must he throw the ball in order to score?
14. A bullet having a muzzle velocity of 150 m/s is fired horizontally from a height of 1.5 m . How much time passes before it strikes the ground? What is its range?
15. A cliff diver in Acapulco launches himself at 1.5 m/s at an angle of 15o above the horizontal from a cliff 60 m above the surf. In order to survive, he must clear the jagged cliffs with a horizontal displacement of 10 m. Will he live to dive again?
THE
END