Quadratic Application: Projectile Motion

Quadratic Applications: Projectile Motion

Generally speaking, projectile motion problems involve objects that are thrown, shot, or dropped. Usually the object will be launched directly upward or dropped directly down.

**Consider the following example:**

An object is launched directly upward at 19.6 m/s from a 58.8-meter tall platform. The equation for the object's height (

*s*) at time (

*t*) seconds after launch is

*s*(

*t*) = -4.9

*t*

^{2}+ 19.6

*t*+ 58.8, where

*s*is in meters. When does the object strike the ground?

**Understand**

This question asks for the time when the object strikes the ground. Another way of thinking of this is, when is the object's height zero (on the ground)? So, set s equal to zero and solve the equation:

**Solve**

0 = -4.9t^{2}+ 19.6t+ 58.8

0 =t^{2}- 4t- 12

0 = (t- 6)(t+ 2)t= 6 ort= -2

Recall that

*t*is a value of time, answering the question "when." It doesn't make sense in this context for*t*to be a negative value, so*t*= -2 is an extraneous solution.The object strikes the ground

**6 seconds**after launch.**Formula of Projectile Motion Problems**

The key features to know in projectile motion problems are the initial height, the initial speed, a

value of the force of gravity, and time. They are related in the formula given below:

*s*(

*t*) = -

*gt*

^{2}+

*v*

_{0}

*t*+

*h*

_{0}

where:

is the height at any particular time (*s**t*) [Note:*s*(*t*) is also sometimes shown in the formula as*h*]is gravity value – in feet this value is 16 and in meters this value is 4.9 [Note: In physics, the gravitational constant is actually 32 for feet and 9.8 for meters, but the formula uses one-half this value.]**g**is the initial velocity*v*_{0}is the initial height*h*_{0}