PROJECTILE MOTION CHAPTER 3
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PROJECTILE MOTIONCHAPTER 3
PROJECTILE MOTION
An object launched into space without motive power of its own is called a projectile.
If we neglect air resistance, the only force acting on a projectile is its weight, which causes its path to deviate from a straight line.
The projectile has a constant horizontal velocity and a vertical velocity that changes uniformly under the influence of the acceleration due to gravity.
HORIZONTAL PROJECTION If an object is projected horizontally, its motion can best be described by considering its horizontal and vertical motion separately.
In the figure we can see that the vertical velocity and position increase with time as those of a free-falling body. Note that the horizontal distance increases linearly with time, indicating a constant horizontal velocity.
3.19 A cannonball is projected horizontally with an initial velocity of 120 m/s from the top of a cliff 250 m above a lake.a. In what time will it strike the water at the foot of the cliff?
b. What is the x-distance (range) from the foot of the cliff to the point of impact in the lake?
v0x = 120 m/s
y = 250 mv0y = 0 g
yt
2
8.9
)250(2 = 7.14 s
x = vx t
= 120(7.14) = 857 m
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c. What are the horizontal and vertical components of its final velocity?
d. What is the final velocity at the point of impact and its direction?
vx = 120 m/s
vy = voy + gt
= 0 + 9.8 (7.14) = 70 m/s
22yxR vvv
22 )70()120(
120
70tan 1
= 139 m/s
= 30.2 below horizontal
v (139 m/s, 30.2)
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3.20 A person standing on a cliff throws a stone with a horizontal velocity of 15.0 m/s and the stone hits the ground 47 m from the base of the cliff. How high is the cliff?
vx = 15 m/s
x = 47 mvy = 0
xv
xt
15
47 = 3.13 s
y = ½ gt2
= ½ (9.8)(3.13)2
= 48 m
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PROJECTILE MOTION AT AN ANGLEThe more general case of projectile motion occurs when the projectile is fired at an angle.
Problem Solution Strategy:
1. Upward direction is positive. Acceleration due to gravity (g) is downward thus g = - 9.8 m/s2
2. Resolve the initial velocity vo into its x and y components:
vox = vo cos θ voy = vo sin θ
3. The horizontal and vertical components of its position at any instant is given by: x = voxt y = voy t +½gt2
4. The horizontal and vertical components of its velocity at any instant are given by: vx = vox vy = voy + gt
5. The final position and velocity can then be obtained from their components.
3.23 An artillery shell is fired with an initial velocity of 100 m/s at an angle of 30 above the horizontal. Find:a. Its position and velocity after 8 s
vo = 100 m/s, 30t = 8 sg = - 9.8 m/s2
vox = 100 cos 30 = 86.6 m/svoy = 100 sin 30 = 50 m/s
x = vox t
= 86.6(8) = 692.8 my = voy t + ½ gt2
= 50(8) + ½ (-9.8)(8)2
= 86.4 m
vx = vox = 86.6 m/s
vy = voy + gt
= 50 + (-9.8)(8) = - 28.4 m/s
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b. The time required to reach its maximum height
At the top of the path: vy = 0
vy = voy + gt g
vt oy
8.9
50
= 5.1 s
c. The horizontal distance (range)
Total time T = 2t = 2(5.1) = 10.2 s
x = vox t
= 86.6(10.2) = 883.7 m
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3.24 A baseball is thrown with an initial velocity of 120 m/s at an angle of 40above the horizontal. How far from the throwing point will the baseball attain its original level?
vo = 120 m/s, 40g = - 9.8 m/s2
vox = 120 cos 40 = 91.9 m/svoy = 120 sin 40 = 77.1 m/sAt top vy = 0
g
vt oy
8.9
1.77
= 7.9 s
x = vox (2t)
= 91.9(2)(7.9) = 1452 m
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3.25 A plastic ball that is released with a velocity of 15 m/s stays in the air for 2.0 s.
a. At what angle with respect to the horizontal was it released?
vo = 15 m/s
t = 2 stime to maximum height = 1 sat the top vy = 0
vy = voy + gt
g
vt o sin
ov
tgsin
15
)1(8.9sin 1 = 40.8º
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b. What was the maximum height achieved by the ball?
y = voy t +½ gt2
= (15)(sin 40.8º)(1) + ½ (-9.8)(1)2
= 4.9 m
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3.26 Find the range of a gun which fires a shell with muzzle velocity vo at an angle θ .
At top vy = 0
vy = voy + gt
= vo sin θ - gt
g
vt o sin
Total time = 2t
g
vv oo
sin2cos
)cos(sin2 2
g
vx o
x = vxt
K2D
sin θ cos θ = ½ sin 2θ
2sin2
g
vx o
)cos(sin2 2
g
vx o
b. What is the angle at which the maximum range is possible?
Maximum range is 45 since 2θ = 90
sin θ cos θ = ½ sin 2θ
2sin2
g
vx o
)cos(sin2 2
g
vx o
b. What is the angle at which the maximum range is possible?
Maximum range is 45 since 2θ = 90
c. Find the angle of elevation of a gun that fires a shell with muzzle velocity of 120 m/s and hits a target on the same level but 1300 m distant.
vo = 120 m/s
x = 1300 m
22sin
ov
gx
2)120(
)1300(8.9
2sin2
g
vx o
= 0.885
sin-1(2θ)= 62
θ = 31
PROJECTILE MOTION
An object launched into space without motive power of its own is called a projectile.
If we neglect air resistance, the only force acting on a projectile is its weight, which causes its path to deviate from a straight line.
c. Find the angle of elevation of a gun that fires a shell with muzzle velocity of 120 m/s and hits a target on the same level but 1300 m distant.
vo = 120 m/s
x = 1300 m
22sin
ov
gx
2)120(
)1300(8.9
2sin2
g
vx o
= 0.885
sin-1(2θ)= 62
θ = 31
c. Find the angle of elevation of a gun that fires a shell with muzzle velocity of 120 m/s and hits a target on the same level but 1300 m distant.
vo = 120 m/s
x = 1300 m
22sin
ov
gx
2)120(
)1300(8.9
2sin2
g
vx o
= 0.885
sin-1(2θ)= 62
θ = 31
PROJECTILE MOTION
An object launched into space without motive power of its own is called a projectile.
If we neglect air resistance, the only force acting on a projectile is its weight, which causes its path to deviate from a straight line.
c. Find the angle of elevation of a gun that fires a shell with muzzle velocity of 120 m/s and hits a target on the same level but 1300 m distant.
vo = 120 m/s
x = 1300 m
22sin
ov
gx
2)120(
)1300(8.9
2sin2
g
vx o
= 0.885
sin-1(2θ)= 62
θ = 31
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