Math Insight

Another differential equation: projectile motion


Here we encounter the fundamental idea that if $s=s(t)$ is position, then $\dot{s}$ is velocity, and $\ddot{s}$ is acceleration. This idea occurs in all basic physical science and engineering.

In particular, for a projectile near the earth's surface travelling straight up and down, ignoring air resistance, acted upon by no other forces but gravity, we have $$\hbox{ acceleration due to gravity }= -32 \hbox{ feet/sec }^2.$$ Thus, letting $s(t)$ be position at time $t$, we have $$\ddot{s}(t)=-32.$$ We take this (approximate) physical fact as our starting point.

From $\ddot{s}=-32$ we integrate (or anti-differentiate) once to undo one of the derivatives, getting back to velocity: $$v(t)=\dot{s}=\dot{s}(t)=-32t+v_o$$ where we are calling the constant of integration ‘$v_o$’. (No matter which constant $v_o$ we might take, the derivative of $-32t+v_o$ with respect to $t$ is $-32$.)

Specifically, when $t=0$, we have $$v(0)=v_o$$ Thus, the constant of integration $v_o$ is initial velocity. And we have this formula for the velocity at any time in terms of initial velocity.

We integrate once more to undo the last derivative, getting back to the position function itself: $$s=s(t)=-16t^2+v_ot+s_o$$ where we are calling the constant of integration ‘$s_o$’. Specifically, when $t=0$, we have $$s(0)=s_o$$ so $s_o$ is initial position. Thus, we have a formula for position at any time in terms of initial position and initial velocity.

Of course, in many problems the data we are given is not just the initial position and initial velocity, but something else, so we have to determine these constants indirectly.


  1. You drop a rock down a deep well, and it takes $10$ seconds to hit the bottom. How deep is it?
  2. You drop a rock down a well, and the rock is going $32$ feet per second when it hits bottom. How deep is the well?
  3. If I throw a ball straight up and it takes $12$ seconds for it to go up and come down, how high did it go?