One of the most useful applications of Calculus in everyday life is finding the rate of change of a function. Finding the Trajectory of a moving object is a direct example of such. This tutorial will help explain how to use basic calculus methods to solve an otherwise overwhelming problem.

Sample Problem

A circus clown is shot out of a cannon to a landing platform across the stage. The distance from the clown to the ground is given by the formula s(t)=12+176t-16t^2 where t is the number of seconds the clown is in the air from the moment it shot, and s(t) is the number of feet above the ground the clown is at t seconds.

height of the clown with respect to time

  1. Find a formula for the velocity function.
  2. At what speed did the clown emerge from the cannon?
  3. How high was the clown above the ground the moment he was fired?
  4. The clown will go up and then down. How many seconds has he been flying when he reaches the highest point of his trajectory?
  5. How high does the clown go?
  6. How long has the clown been in the air when he lands?


  1. Since s(t) gives us our position at time t, the change in position would give us our velocity or speed. The change in a function is given by its derivative so to find the velocity function we simply take the derivative of the position function. Using the power rule for derivatives we know that v(t)=s^\prime(t)=176-32t
  2. Since we have the velocity function now, we can simply plug in a 0 for time to get the speed at the moment of firing and get v(0)=176-32(0)=176 feet/sec.
  3. Likewise, since we know the position function, we can plug in a 0 for time and get an initial position of s(0)=12+176(0)-16(0)^2=12 feet.
  4. Since he is moving against gravity, there is a point at which he achieves maximum height at a speed of 0. Knowing this, we can set the velocity function equal to 0 and solve for t to find the time at which this occurs. 0=176-32t Using some basic algebra we find that t=176/32=5.5 so 5.5 seconds is our answer.
  5. Now that we have figured out the time at which the clown achieves a max peak, we can plug the time into the position function to learn the height. s(5.5)=12+176(5.5)-16(5.5)^2=496 feet.
  6. When the clown lands the position will be 0. Setting the position function to 0 gives us 0=12+176t-16t^2. Since this is a quadratic equation we need to either factor this or use the quadratic formula to find our values. This particular equation does not factor so we will use the quadratic formula \frac{-b\pm\sqrt{b^2-4ac}}{2a}. Plugging in -16 for a, 176 for b, and 12 for c we get \frac{-176\pm\sqrt{176^2-4(-16)(12)}}{2(-16)}=-5.5\pm\sqrt{31} which can be represented as approximately 11.068 and -.068. We can’t have a negative time so 11.068 seconds is our answer.

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