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Physics Capsule

Projectile motion

Classical mechanics

Projectile motion

From your car keys tossed across the room to missiles launched across the world, projectiles are everywhere. We study the physics driving them.

We have previously seen Newton’s three laws of motion (first, second, and third), and we have also derived the equations of rectilinear motion. We can use this knowledge to understand how a more real-world phenomenon works: projectile motion.

The two axes of projectile motion, shown in blue and orange.

The two axes of projectile motion, shown in blue and orange.

While the term “projectile motion” may sound like a complex system involving projectiles is studied, the reality is much simpler. In physics, a “projectile” is anything which is launched, or thrown, in to the air and covers a certain distance before falling back to the earth under the influence of gravity. So, while a projectile can be something such as a missile launched with some thruster system, a tennis ball you throw is as much a projectile as that missile. Before we begin to understand the mathematics that governs the motion of such projectiles, here is a quick recap of the equations of rectilinear (or one-dimensional) motion that we will be using to study projectile motion which is two dimensional (the projectile goes away from the thrower as well as up and down, giving us two dimensions as shown by the generic projectile motion graph above).

The equation \vec{v}=\vec{v_0}+ \vec{a} t shows how the velocity \vec{v} of an object at any time t is related to its initial velocity, \vec{v_0}, and its constant acceleration, \vec{a}. A similar equation may be written in terms of position and velocity as \vec{x}=\vec{x_0} + \vec{v_{avg}} t, where \vec{v_{avg}} is the average velocity over time t=0, when \vec{x} = 0 = \vec{x_0}, to some time t when the particle is at position \vec{x}. If the average velocity is identified as \vec{v_{avg}} = \frac{\vec{v} + \vec{v_0}}{2}, then \vec{x} - \vec{x_0} = \frac{\vec{v} + \vec{v_0}}{2} t, or \vec{x} - \vec{x_0} = \vec{v_0} t + \frac{1}{2} at^2.

Position and velocity parameters

It helps to get an overview of how we will be solving this problem. The following is our plan of attack, which you can follow using the diagram below: a projectile moves along the x-axis and along the y-axis, which gives it a maximum height at some point and a maximum range as well. In other words, we have four parameters to calculate to understand how projectiles work and their exact spatial position at any point in time. This is quintessential classical mechanics: attempting to predict and map the behaviour of objects over space and/or time.

Position co-ordinates, height, and range.

Position co-ordinates, height, and range.

The simplest parameter to compute would be the x co-ordinate because it has no acceleration along its direction. That is to say, a projectile does not accelerate forwards in the direction of its range. Consider the equation discussed above with \vec{a} = 0, i.e. \vec{x} - \vec{x_0} = \vec{v_0} t. Notice, however that the initial velocity, \vec{v_0} was not in the x-direction, but was at some angle, say \theta with respect to the ground. Therefore, the x-component of \vec{v_0} becomes \vec{v_0} \cos \theta, giving us \vec{x} - \vec{x_0} = ( \vec{v_0} \cos \theta) t. This is our first projectile equation, which, at any time t tells us how far the projectile has moved away along the x-axis.

So what about the y-component that describes how high the projectile is at any given time t? This situation is similar to how a body behaves during free fall because gravity is essentially the only force acting on the projectile influencing its y-directional movement. (I say essentially because air resistance too exists but is often small enough that it maybe neglected; it does not have to be small, but that case is best left for another day.) For an object in free fall, the same equations of rectilinear motion apply except they are all under the influence of a constant acceleration due to the earth’s gravity, \vec{g}, taken to be roughly 9.8 ms^{-2}.

Therefore, the same equation for \vec{x}-\vec{x_0} in the y-component form would be \vec{y}-\vec{y_0} = \vec{v_0}^{\prime} t - \frac{1}{2} \vec{g} t^2, or, using the component form for \vec{v_0}^{\prime}, we get \vec{y}-\vec{y_0} = ( \vec{v_0} \sin \theta ) t - \frac{1}{2} \vec{g} t^2.

Consequently, the velocity in the y-direction \vec{v_y} = \vec{v_0}^{\prime} - \vec{g}t for \vec{g} considered negative by convention, and since we can use the y-component with \sin \theta as we did before, we arrive at \vec{v_y} =  \vec{v_0} \sin \theta - \vec{g}t. And with this we now know the exact position of the projectile at any given time as well as how fast it is rising or falling — barring any small errors due to air resistance, of course.

Path and range measurement

To sum up what we have seen so far, in studying projectiles our interest lies in four areas: we want to be able to locate the projectile exactly at any point of time, we want to know how fast it is rising or falling, we want to know how it travels (Is it straight? Is it circular?), and we want to know how far it can go. We have, so far, covered the first two. Next, then, is all about the path that the projectile traces.

It turns out that this path is actually easy to calculate. We have already seen that \vec{x} - \vec{x_0} = ( \vec{v_0} \cos \theta) t, or alternatively,

    \begin{align*} t = \frac{\vec{x} - \vec{x_0}}{( \vec{v_0} \cos \theta)} \end{align*}

Substitute this for the y-positional co-ordinate, \vec{y}-\vec{y_0} = ( v_{0} \sin \theta ) t - \frac{1}{2} \vec{g} t^2, and we arrive at,

    \begin{align*} \vec{y}-\vec{y_0} = ( \vec{v_0} \sin \theta ) \frac{\vec{x} - \vec{x_0}}{( \vec{v_0} \cos \theta)} - \frac{1}{2} \vec{g} \left( \frac{\vec{x} - \vec{x_0}}{( \vec{v_0} \cos \theta)} \right) ^2 \end{align*}

Therefore, at \vec{x_0} = 0 and \vec{y_0} = 0, we finally get,

    \begin{align*} \vec{y} = (\tan \theta ) x - \frac{\vec{g} \vec{x}^2}{2 (\vec{v_0} \cos \theta )^2} \end{align*}

Notice how the equation looks like y = ax + bx^2, which is the equation for a parabola. This simply tells us that all projectiles follow a path that traces out a parabola.

So just how far does this parabolic path take our projectile. The simplest answer would be, “till it reaches the ground again” but it could hit a bird, a plane, or superman, so let us brush aside all these particular situations and consider the ideal one: what is the maximum distance a parabola can cover under given circumstances of initial velocity, \vec{v_0}, and launch angle, \theta? Looking at this another way, we ask, “what is the maximum value of x under the given circumstances?”

Our condition, understandably, is that y = y_0, or the initial and final heights are the same, i.e. a tennis ball somehow thrown from, say, the mean sea level returns to the mean sea level. Let us also define x - x_0 as the range, R. Therefore,

    \begin{align*} x - x_0 = R = ( v_0 \cos \theta) t \end{align*}

and,

    \begin{align*} y - y_0 = ( v_0 \sin \theta ) t - \frac{1}{2} g t^2 = 0 \end{align*}

We can solve these two equations to arrive at,

    \begin{align*} t &= \frac{R}{( v_0 \cos \theta)} \\ \Rightarrow 0 &= ( v_0 \sin \theta ) - \frac{1}{2} g \frac{R}{( v_0 \cos \theta)} \\ \Rightarrow R &= \frac{2 v_0^2 \sin \theta \cos \theta}{g} \end{align*}

Since we have the mathematical identity, \sin 2\theta = 2 \sin \theta \cos \theta,

    \begin{align*} R = \frac{v_0^2}{g} \sin 2 \theta \end{align*}

This is the range in, understandably, the ideal case. For y - y_0 \neq 0, necessary changes have to be made. Note that the height, y, is maximum at x = \frac{R}{2}. Also note that the range is maximum for a given initial velocity when \sin 2 \theta is maximum, or 1. This occurs for 2 \theta = 90^{o} or \theta = 45^{o}, once again, in an ideal case without air resistance etc. And, as I write this, is dawns on me how a lot of physics really is about spherical chickens in a vacuum.

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V.H. Belvadi is an Assistant Professor of Physics. He teaches postgraduate courses in advanced classical mechanics, astrophysics and general relativity. When he is free he makes photographs and short films, writes on his personal website, makes music, reads voraciously, or plays his violin. He currently serves as the Editor-in-Chief of Physics Capsule.

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