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Given these assumptions, the following steps are then used to analyze projectile motion:
Step 1
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These axes are perpendicular, so Ax=Acosθ and Ay=Asinθ are used
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The
magnitudes of the components of the velocity v are vx=vcosθ and vy=vsinθ, where v is the
magnitude of the velocity and θ is its direction, as shown in Figure 3
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Initial values are denoted
with a subscript 0, as usual
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Treat the motion as two independent one-dimensional motions, one horizontal and the
other vertical
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Solve for the unknowns in the two separate motions—one horizontal and one vertical
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The problem-solving
procedures here are the same as for one-dimensional kinematics and are illustrated in the solved
examples below
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Recombine the two motions to find the total displacement s and velocity v
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Galileo was the first person to fully
comprehend this characteristic
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On level ground,
we define range to be the horizontal distance R traveled by a projectile
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However, investigating the range of projectiles can shed light on other interesting
phenomena, such as the orbits of satellites around the Earth
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How does the initial velocity of a projectile affect its range? Obviously, the greater the
initial speed v0, the greater the range, as shown in Figure 3
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The initial angle θ0 also has a
dramatic effect on the range, as illustrated in Figure 3
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For a fixed initial speed, such as
might be produced by a cannon, the maximum range is obtained with θ0=45º
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If air resistance is considered, the maximum angle is
approximately 38º
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The range also depends on the value of the
acceleration of gravity g
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The range R of a projectile on level ground
for which air resistance is negligible is given by

where v0 is the initial speed and θ0 is the initial angle relative to the horizontal
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When we speak of the range of a projectile on level ground, we assume that R is very
small compared with the circumference of the Earth
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The range is larger than predicted by the range equation given above because the projectile has
farther to fall than it would on level ground
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41
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This possibility was recognized centuries before it could be
accomplished
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The object thus falls continuously but never hits the surface
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Once again, we see that thinking about one topic, such as the range of a projectile, can
lead us to others, such as the Earth orbits
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DEFINITION OF TERMS
Air Resistance -a frictional force that slows the motion of objects as they travel through the air;
when solving basic physics problems, air resistance is assumed to be zero
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Classical Relativity -the study of relative velocities in situations where speeds are less than about
1% of the speed of light—that is, less than 3000 km/s
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Component (Of A 2-D Vector) -a piece of a vector that points in either the vertical or the
horizontal direction; every 2-d vector can be expressed as a sum of two vertical and horizontal
vector components
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Head (Of A Vector) -the end point of a vector; the location of the tip of the vector’s arrowhead;
also referred to as the “tip”
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Kinematics -the study of motion without regard to mass or force
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Motion -displacement of an object as a function of time
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Projectile Motion -the motion of an object that is subject only to the acceleration of gravity
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Relative Velocity -the velocity of an object as observed from a particular reference frame
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Resultant -the sum of two or more vectors
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Scalar -a quantity with magnitude but no direction
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Trajectory -the path of a projectile through the air
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Vector Addition -the rules that apply to adding vectors together
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PROBLEM SET
Problem 1
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Time elapsed during the motion is 5s, calculate the height that object is thrown and Vy
component of the velocity after it hits the ground
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John kicks the ball and ball does projectile motion with an angle of 53º to horizontal
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(sin53º=0, 8 and cos53º=0, 6)

Problem 3
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Find the maximum height it
can reach, horizontal distance it covers and total time from the given information
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You are trapped on the top of a burning building
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There is a safe building 6
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00 m lower
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You run horizontally off your building at 8
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Do you make it across? If you don't,
how much faster must you be going?
First we sketch the situation and possible outcomes
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We solve projectile motion problems by considering
the x and y components separately, keeping in mind that the time in air is common
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Looking at the x information, we see that we have enough data to find 𝑡𝑎𝑖𝑟
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Since ax = 0 for a projectile, this
equation become Δx = v0x t
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50 m
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On the other hand,
looking at the y information, we see that we also have enough data to find 𝑡𝑎𝑖𝑟
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We know v0y = 0 since you ran off
the roof horizontally and that ay = g, thus this equation becomes Δy = ½gt 2
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00 m
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Since the time it takes to cross the horizontal distance is less than the time you have, you have
don't make it across
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Since ax = 0 for
projectiles, the kinematics equation become Δx = 𝑉𝑜𝑥 t where t is now the 0
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Solving for
𝑉𝑜𝑥 , we get ,
If you were able to run at 8
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Problem 5
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At the height of 2 km, an object
is dropped from the aircraft
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Known :
Height = 2 km = 2000 meters
Acceleration due to gravity (g) = 10 m/s2
Wanted : The time interval (t)
Solution :
h = 1/2 g t2
2000 = 1/2 (10) t2
2000 = 5 t2
t2 = 2000/5 = 400
t = √400 = 20 seconds

Problem 6
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Determine the distance of X
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Known :
Initial speed (vo) = 25 m/s
Acceleration due to gravity (g) = 10 m/s2
Angle (θ) = 45o
Wanted : X
Solution :
The horizontal component of the initial velocity :
vox = vo cos θ = (25 m/s)(cos 45o) = (25 m/s)(0
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5√2 m/s
The vertical component of the initial velocity :
𝑣𝑜𝑦 = vo sin θ = (25 m/s)(sin 45o) = (25 m/s)(0
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5√2 m/s
Projectile motion could be understood by analyzing the horizontal and vertical component of the
motion separately
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Time in the air (t) :
The time in air calculated with the equation of the upward vertical motion
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Known :
The initial velocity (vo) = 12
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5√2) t + 1/2 (-10) t2
0 = 12
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5√2 t = 5 t2
12
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5√2 / 5
t = 2
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Known :
Velocity (v) = 12
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5√2 seconds
Wanted : Distance
Solution :
d = v t = (12
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5√2) = (12
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5)(2) = 62
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An object projected upward at an angle θ = 30𝑜 with the horizontal has an initial speed of 20 m/s
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Determine the maximum height
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5) = 10 m/s
Calculate the maximum height
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Known :
Acceleration due to gravity (g) = -10 m/s2 (downward direction, negative)
The vertical component of the initial velocity (voy) = 10 m/s (upward direction, positive)
Velocity at the maximum height (vty) = 0
Wanted : The maximum height (h)
Solution :
vt2 = vo2 + 2 g h
02 = 102 + 2 (-10) h
0 = 100 – 20 h
100 = 20 h
h = 100/20
h = 5 meters
The maximum height is 5 meters
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A balloon having 20 m/s constant velocity is rising from ground to up
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Find the horizontal distance travelled by the object
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An object hits the ground as given in the picture below
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Problem 10
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If the initial velocity of the object
is 50m/s, find the time of motion, maximum height it can reach, and distance in horizontal
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A bullet is fired at an initial velocity of 150 m/s and an angle of 56° at the top of a 120 m
tall building
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The time for the bullet to hit the ground
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Problem 12
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How fast must the basketball player shoot the ball so that it goes
through the hoop without touching the backboard?

Problem 13
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0m/s
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What is the angle below the horizontal of the velocity of the
puck just before it hits the ground?

Problem 14
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Gravity =
9
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What is the maximum height the projectile reaches?

Let’s set up what we know
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𝑣0 = initial velocity = muzzle velocity = 150 m/s
𝑣𝑥 = horizontal velocity component
𝑣𝑦 = vertical velocity component
θ = angle of elevation = 45°
h = maximum height
R = range
x = horizontal position at t=10 s
y = vertical position at t=10 s
m = mass of projectile
g = acceleration due to gravity = 9
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1 m/s
F = ma = -mg
solve for a
a = -g

vhy – v0y = at
0 – v0y = -9
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1 m/s = -9
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8 s
Now solve the first equation for h
h = v0yt + ½at2
h = (106
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8 s) + ½(-9
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8 s)2
h = 1145
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5 m
h = 574
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4 meters
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A cannon is fired with muzzle velocity of 150 m/s at an angle of elevation = 45°
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8 m/𝑠 2
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First, let’s define our variables
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8 m/𝑠 2
ttotal = tup + tdown
The same acceleration force acts on the projectile in both directions
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tup = tdown
or
ttotal = 2 tup
we found tup in Part a of the problem: 10
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8 s)
ttotal = 21
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6 seconds
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A bullet fired at an angle θ = 60o with a velocity of 20 m/s
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What is the time interval to reach the maximum height?
2
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The height of the object is the same when the time
interval = 1 second and 3 seconds
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