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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.
Practice Key Terms 7

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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