Related Rates: Rocket

Question

A rocket is launched from a platform at an initial speed of 2,000 ft/s. The platform is 100 feet off the ground, and the acceleration due to gravity is -16 ft/s^2. Find the height of the rocket after 3 seconds.

Answer 1

When talking about trajectories or the curved pathway of an object that is shot into the air and pulled down by gravity, we are really just applying a quadratic equation to a real-life situation.

Quadratic equations are written in the standard form of y = a*x² + b*x + c, where a determines the vertical stretch of the parabola and c is the y-intercept. Quadratic equations can also be written in vertex form, but that won't be helpful here, so I won't go into it.

When using a quadratic equation to describe a trajectory, the equation has the form of:

h = (acceleration of the object) * t² + (initial velocity of the object) * t + (initial height of the object)

In this formula, h is the height of the object at any given moment, t.

We are told that the initial height of the rocket is 100 ft and that its initial velocity is 2000 ft/sec. The rocket does not propel itself and so it's susceptible to the force of gravity, which confers an acceleration on the rocket of -16 ft/sec². Acceleration is negative because the force of gravity pulls the rocket down, so the rocket accelerates downward towards the ground even though it's initial velocity was directed upward.

Plugging in these values into our quadratic formula, we get:

h = -16 * t² + 2000 * t + 100

Before going on, let's make sure this equation makes sense... Just for a second, imagine the trajectory of a rocket shot into the air from a raised platform: the path of the rocket would look like an upside-down "U" or an upside parabola. When the leading coefficient in a quadratic equation is positive, the parabola looks like an upright "U", but when the leading coefficient is negative, the parabola looks like an upside-down "U". The leading coefficient in our equation should be negative (and it is) since our trajectory is an upside-down parabola.

To finish this problem, we only have to substitute 3 sec in for t and then solve for h:

h = -16 * t² + 2000 * t + 100

h = -16 * (3)² + 2000 * (3) + 100

h = -144 + 6000 + 100

h = 5956 ft

That wraps it up. Let me know if you still have any questions.