Word Lesson: Quadratic Max/Min Application - Projectiles

• know how to solve quadratic equations:
• know vertex formula for a parabola
• write and solve an equation for the problem

To analyze our problems, we will be using a formula for a freely falling body in which we can ignore any effects of air resistance.

 

 

• s(t) represents the projectile's instantaneous height at any time t
• v_o represents initial velocity
• s_o represents the initial height from which the projectile is released
• t represents time in seconds after the projectile is released

 

In this formula, -16 is a constant which represents one-half of the acceleration due to gravity on the "surface" of the earth (-32 ft/sec²).

In physics, the variable g represents the gravitational field strength (g = F/m) on the surface of the earth. This property is measured in an unusual unit: lbs/slug where lbs measures the gravitational force between an object and the Earth based on the object's mass, the mass of the Earth, and the distance between them; a slug is the English system's unit for mass. You can see this unit's common use when you say that a massive object acts "sluggishly." Through a series of conversions, the unit lbs/slug can be shown to be equivalent to our unit ft/sec².

Since this constant is measured in ft/sec², we must also measure s(t), v_o , and s_o in terms of feet and seconds.

 

Let's begin by substituting known values for variables in the formula:

 

 

Since the formula represents a parabola, we must find the vertex of the parabola to find the time it takes for the ball to reach its maximum height as well as the maximum height (called the apex) . Using the vertex formula:

 

seconds

 

Substituting into the projectile motion formula we have:

 

feet

 

Therefore, if a ball is thrown directly upward from an initial height of 200 feet with an initial velocity of 96 feet per second, after 3 seconds it will reach a maximum height of 344 feet.