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  • Identify and explain the properties of a projectile, such as acceleration due to gravity, range, maximum height, and trajectory.
  • Determine the location and velocity of a projectile at different points in its trajectory.
  • Apply the principle of independence of motion to solve projectile motion problems.

Projectile motion is the motion    of an object thrown or projected into the air, subject to only the acceleration of gravity. The object is called a projectile    , and its path is called its trajectory    . The motion of falling objects, as covered in Problem-Solving Basics for One-Dimensional Kinematics , is a simple one-dimensional type of projectile motion in which there is no horizontal movement. In this section, we consider two-dimensional projectile motion, such as that of a football or other object for which air resistance     is negligible .

The most important fact to remember here is that motions along perpendicular axes are independent and thus can be analyzed separately. This fact was discussed in Kinematics in Two Dimensions: An Introduction , where vertical and horizontal motions were seen to be independent. The key to analyzing two-dimensional projectile motion is to break it into two motions, one along the horizontal axis and the other along the vertical. (This choice of axes is the most sensible, because acceleration due to gravity is vertical—thus, there will be no acceleration along the horizontal axis when air resistance is negligible.) As is customary, we call the horizontal axis the x -axis and the vertical axis the y -axis. [link] illustrates the notation for displacement, where s size 12{s} {} is defined to be the total displacement and x size 12{x} {} and y size 12{y} {} are its components along the horizontal and vertical axes, respectively. The magnitudes of these vectors are s , x , and y . (Note that in the last section we used the notation A size 12{A} {} to represent a vector with components A x size 12{A rSub { size 8{x} } } {} and A y size 12{A rSub { size 8{y} } } {} . If we continued this format, we would call displacement s size 12{s} {} with components s x size 12{s rSub { size 8{x} } } {} and s y size 12{s rSub { size 8{y} } } {} . However, to simplify the notation, we will simply represent the component vectors as x size 12{x} {} and y size 12{y} {} .)

Of course, to describe motion we must deal with velocity and acceleration, as well as with displacement. We must find their components along the x - and y -axes, too. We will assume all forces except gravity (such as air resistance and friction, for example) are negligible. The components of acceleration are then very simple: a y = g = 9.80 m /s 2 size 12{a rSub { size 8{y} } ="-g"="-9.80" "m/s" rSup { size 8{2} } } {} . (Note that this definition assumes that the upwards direction is defined as the positive direction. If you arrange the coordinate system instead such that the downwards direction is positive, then acceleration due to gravity takes a positive value.) Because gravity is vertical, a x = 0 size 12{a rSub { size 8{x} } } {} . Both accelerations are constant, so the kinematic equations can be used.

Review of kinematic equations (constant a )

x = x 0 + v - t size 12{x=`x rSub { size 8{0} } `+` { bar {v}}t} {}
v - = v 0 + v 2 size 12{ { bar {v}}=` { {v rSub { size 8{0} } +v} over {2} } } {}
v = v 0 + at size 12{v=v rSub { size 8{0} } + ital "at"} {}
x = x 0 + v 0 t + 1 2 at 2 size 12{x=x rSub { size 8{0} } +v rSub { size 8{0} } t+ { {1} over {2} } ital "at" rSup { size 8{2} } } {}
v 2 = v 0 2 + 2 a ( x x 0 ) . size 12{v rSup { size 8{2} } =v rSub { size 8{0} } rSup { size 8{2} } +2a \( x - x rSub { size 8{0} } \) } {}
A soccer player is kicking a soccer ball. The ball travels in a projectile motion and reaches a point whose vertical distance is y and horizontal distance is x. The displacement between the kicking point and the final point is s. The angle made by this displacement vector with x axis is theta.
The total displacement s size 12{s} {} of a soccer ball at a point along its path. The vector s size 12{s} {} has components x size 12{x} {} and y size 12{y} {} along the horizontal and vertical axes. Its magnitude is s size 12{s} {} , and it makes an angle θ size 12{θ} {} with the horizontal.

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
Aislinn Reply
cm
tijani
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John Reply
what is physics
Siyaka Reply
A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
Jude Reply
Can you compute that for me. Ty
Jude
what is the dimension formula of energy?
David Reply
what is viscosity?
David
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emma Reply
what is chemistry
Youesf Reply
what is inorganic
emma
Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes
Adjei
please, I'm a physics student and I need help in physics
Adjanou
chemistry could also be understood like the sexual attraction/repulsion of the male and female elements. the reaction varies depending on the energy differences of each given gender. + masculine -female.
Pedro
A ball is thrown straight up.it passes a 2.0m high window 7.50 m off the ground on it path up and takes 1.30 s to go past the window.what was the ball initial velocity
Krampah Reply
2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
Sahid Reply
you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
Ryan
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Maurice Reply
what are the types of wave
Maurice
answer
Magreth
progressive wave
Magreth
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Muhammad Reply
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Mohammed
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Mujahid
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
yasuo Reply
Who can show me the full solution in this problem?
Reofrir Reply
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Source:  OpenStax, Physics 110 at une. OpenStax CNX. Aug 29, 2013 Download for free at http://legacy.cnx.org/content/col11566/1.1
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