Mechanics: simple projectile motion – 1


Since I’ve been playing with simple projectile motion recently, I thought I would write it up and post it.

In fact, there will be two posts: the first will not require calculus, but the second post will.

This is far from high-powered mathematics… but I enjoyed it.

We do this modeling in a very simple universe:

  • the Earth is flat;
  • gravity is constant;
  • there is no atmosphere (to be specific, no air resistance);
  • and the earth doesn’t rotate (it’s an inertial system, and the ground isn’t moving while the projectile is in the air).

My coordinate system is the usual xy-plane: x is horizontal, y is vertical, the acceleration of gravity is in the negative y direction.

I take the following four equations as given. (This is what I did in high school, before I had had calculus. The derivations are almost trivial, if you’ve had calculus, but I’ll try to remember to show them in the next post.)

\text{sx}(t)=\text{sxo}+t\ \text{vxo}

\text{sy}(t)=-\frac{g\ t^2}{2}+\text{syo}+t\ \text{vyo}


\text{vy}(t)=\text{vyo}-g\ t


  • an initial s denotes distance
  • an initial v denotes velocity
  • a final o denotes an initial value, i.e. at time t = 0.
  • g is the acceleration of gravity
  • t is time

that is:

  • sx[t] is the x-component (horizontal) of distance as a function of time t.
  • sy[t] is the y-component (vertical) of distance as a function of time.
  • sxo is the initial value (i.e. at t = 0) of the x-component of distance, i.e. sxo = sx[0].
  • syo is the initial value of the y-component of distance, i.e. syo = sy[0].
  • vx[t] is the x-component of velocity as a function of time.

  • vy[t] is the y-component of velocity as a function of time.
  • vxo is the initial value of the x-component of velocity, i.e. vxo = vx[0].

  • vyo is the initial value of the y-component of velocity, i.e. vyo = vy[0].

Nevertheless, if at all possible, I will choose my origin (for space and for time) so that

sxo = syo = 0.

For example, if a ball rolls off a table which is 4 feet above the ground, I will chose the edge of the table as the xy-origin, and t = 0 when it rolls off, and then the ground is at

sy = -4 Feet.

Computationally, that means that I usually apply the following rule

BC = {sxo -> 0, syo -> 0}

to the fundamental equations (fe2), which simplifies all of them.

\text{sx}(t)=t\ \text{vxo}

\text{sy}(t)=t\ \text{vyo}-\frac{g\ t^2}{2}


\text{vy}(t)=\text{vyo}-g\ t

Sometimes we will find it necessary to use the rule

BC2 = {vxo -> vo Cos[\alpha],vyo -> vo Sin[\alpha]}

to replace the two initial components vxo and vyo by

{vo Cos[\alpha], vo Sin[\alpha]}

in the four equations. (That is, we’ve been given the initial magnitude and direction.) I should remark that the initial angle \alpha determines the shape of the trajectory, while the initial speed vo determines the scale — the size — of the trajectory. Once in a while we may encounter a problem that depends only on the shape and not on the absolute size of the trajectory.

Oh, we will also need the value of g… and we will need it in both SI and English units. As with all the other numbers, I use rules:

cs= g -> (9.80665 Meter)/Second^2

cs2 = g -> (32.174 Feet)/Second^2

So let’s work that 4-foot high table.

A block slides off a table

(6.2; Sears & Zemansky; College Physics, 3rd ed. 1960.)

A block slides off a horizontal table top 4 feet high with a velocity of 4 ft./s. Find:

  • the horizontal distance from the table at which the block strikes the floor;
  • the components of velocity when it reaches the floor.
  • As is my custom, the table edge is the origin… and let T be the (unknown) time of impact.

    That is, the parameters of this problem are:

    ps = {vyo -> 0, vxo -> (4 Feet)/Second, sy[T] -> -4 Feet}

    Here we have the general equations (that’s the “fe2”)… followed by the equations for this problem at an arbitrary time t… followed by the equations for this problem at time T.

    The first equation says sx is proportional to time…
    the second equation has only T as an unknown, when the block has dropped 4 feet (i.e. it has hit the floor)…
    the third equation says vx is constant…
    the fourth equation says vy is proportional to time.

    Solve the second equation for T, the time to impact…

    Now, plug that back into all the equations at time T…

    which tells us that it hit 2 feet over (okay, 1.9948 feet), and we can read off the components of velocity at impact. That is, for example, the fourth equation says that vy(impact time) = 16.0435 feet/second.

    We’re done.

    Because I worked with all the equations, I got all the answers as soon as I plugged in the time to impact. Mathematica® makes this so convenient.

    I should point out that if I had just asked Mathematica to solve the entire set, it would have had trouble, because T is an argument in vx[T] and v]T]. I could get around that just by renaming those terms, but it isn’t worth it to me.

    (That is, I could replace vx[T] by “vxT” and vy[T] by “vyT”, or some such symbols — just get rid of the functional dependence. I prefer what I did instead.)


    (Example 1; Neville de Mestre, The Mathematics of Projectiles in Sport, 1990.)

    A player hits a baseball into the outfield against a wall 100 m away. if it leaves the bat at an angle of 45{}^{\circ}\ to the horizontal and strikes the wall 10 m above the bat-ball contact position, what is the initial speed of the ball?

    Okay, let T be the time of impact on the wall. The parameters for this problem are

    {sy[T] -> 10 Meter, \alpha -> 45{}^{\circ}\ , sx[T] -> 100 Meter}

    As usual, I take the bat-ball contact position as time zero, and as the origin. In addition, I need to replace vxo and vyo by vo and trig functions of the initial angle.

    BC2 = {vxo -> vo Cos[\alpha], vyo -> vo Sin[\alpha]}

    Here we have the general equations… followed by the equations for this problem at an arbitrary time t… followed by the equations for this problem at time T.

    (Those square roots of 2 are the sine or cosine of 45{}^{\circ}\ .)

    Looking at the last set of equations… the first two describe the final distances, horizontal and vertical respectively. And the first two equations have two unkowns, speed vo and impact time T.

    This time, I am going to solve for T and vo simultaneously. The first two equations are

    … and the solution is:

    {vo -> (33.0095 Meter)/Second, T -> 4.28426 Second}

    (There were two solutions, take the one with positive speed and time. The other solution had the same magnitudes but negative signs; it corresponds to the ball leaving the wall at a negative time and travelling leftwards to the bat, arriving at t = 0.)

    At this point, we’ve answered the question: the initial speed was 33 meter/second.

    But we might as well plug into the full set of equations at time T:

    We can read off the components of velocity as the ball strikes the outfield wall. It’s nice to see that the vertical component is negative, since the ball is falling.
    Now, there are two other things I want to do for this problem. Maybe I should have done them for the first one, too.

    First I want to plot sx,sy as a function of t from t = 0 to t = T. Here are the equations:

    After I drop units, I am left with a 2D curve:

    {sx(t), sy(t)} = {23.3412 t, 23.3412 t – 4.90333 t^2}

    As a quick check of our work, we want to find the maximum altitude of the ball.

    In the first equation below, I set the (specific) potential energy equal to the (specific) kinetic energy. (Because the origin is the initial position, the PE is zero initially, and the KE of the y-component of the velocity is zero at maximum altitude.)

    In the second equation, I have set the parameters; the only unknown is the height h. I solve for it — and then throw away the units so I can graph it.

    When i plot that curve, from t = 0 to T… and also draw a horizontal line at the maxximum altitude, I get…

    Beautiful, if I say so myself. The ball moves from the origin, ending at 100 meters to the right and 10 meters up. (I plotted it as a function of time, so I’m getting a graphical check.) My independent calculation of the maximum altitude (the horizontal line) is consistent with the computed trajectory of the ball.

    a rock falls past a window

    (I made up the numbers, but I remember this problem from high school. It was unforgettable.)

    We are looking out a 4 foot tall window. At t = 0 a stone passes the top (outside). 1/10 second later it passes the bottom. (No, I have no idea how we know that time.)

    Assuming that it was dropped from above, rather than thrown down, from how far up was it dropped?

    As usual, t = 0 is when I first saw it (the top of the window), and that’s also the origin. This is a one-dimensional problem… the stone was dropped from a positve height, but at a negative time.

    The given conditions are

    ps = {T -> Second/10, sy[t] -> -4 Feet, sxo -> 0. Feet, vxo -> (0. Feet)/Second}

    Here we have the general equations… followed by the equations for this problem at an arbitrary time t… followed by the equations for this problem at time T.

    As we would have expected, two of the equations say that sx(t) = vx(t) = 0. The two nontrivial equations are…

    NOTE that vyo is not 0 — when we first see the ball, it is moving. What we are trying to find is the height when the vertical speed was zero, i.e. when vy = 0.

    We have two unknowns, vyo and vy[1/10], so solve. Note that we needed g in English units. And Mathematica won’t be troubled by the argument 1/10 in vy[1/10]. It’s a symbolic T as an argument when I’m trying to find T that would cause difficulties.

    Anyway, the solution is

    {{vy[Second/10] -> -((41.6087 Feet)/Second), vyo -> -((38.3913 Feet)/Second)}}

    Nice, both velocities are negative; the stone is, after all, falling.

    Now we need to find the height when vy(T) = 0. This is the same calculation as the maximum height of the baseball in the previous example.

    That is, conservation of energy. The initial KE came from PE. Take the KE at t = 0 — since that’s the origin, measure PE from the top of the window.

    Call that 23 feet above us. Figuring 15 feet to each story, that’s a story and a half… I guess we’re on the second highest floor, and someone dropped the stone off the roof. (15 feet to the window above me, then 8 feet to the roof.)

    Can I find a way to check this?

    The average speed of the stone while I see it… it travels 4 feet in 1/10 second = 40 fps. How long did it take to get up to that speed? Note that I am moving the origin from the top of the window up to the drop point…

    How far has it travelled in that elapsed time? The origin is still at the drop point:

    It has fallen 24.8648 feet. Guess what? That’s just about two feet more than the height I computed (2 + 22.905 =24.905 feet) … i.e. to the midpoint of the window!

    The discrepancy is that because the speed is not constant, the average speed does not occur at the midpoint of the window, i.e. at half the distance fallen — but at half the time travelled, i.e. at 1/5 second after it passes the top of the window.

    I think our work here is done.


    Leave a Reply

    Fill in your details below or click an icon to log in: Logo

    You are commenting using your account. Log Out /  Change )

    Google+ photo

    You are commenting using your Google+ account. Log Out /  Change )

    Twitter picture

    You are commenting using your Twitter account. Log Out /  Change )

    Facebook photo

    You are commenting using your Facebook account. Log Out /  Change )


    Connecting to %s

    %d bloggers like this: