SOLUTION: An object is thrown vertically upward from the ground with a velocity of 96 ft/s. Its height, y, is measured in feet, after t seconds is given by y = 96t - 16t^2. How high will the

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: An object is thrown vertically upward from the ground with a velocity of 96 ft/s. Its height, y, is measured in feet, after t seconds is given by y = 96t - 16t^2. How high will the      Log On


   

Question 1125211: An object is thrown vertically upward from the ground with a velocity of 96 ft/s. Its height, y, is measured in feet, after t seconds is given by y = 96t - 16t^2. How high will the object be after two seconds? After how many seconds will the object reach its maximum height? What is the maximum height?
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


1. Substitute 2 for and do the arithmetic.

2. The graph of is a parabola that opens downward because the lead coefficient, -16, is less than zero. Therefore the maximum value of will be at the vertex of the parabola. The vertex occurs at the point calculated by dividing the opposite of the linear term by twice the lead coefficient. So for



The vertex is at:



You need the abscissa of this point, , to answer "After how many seconds..." and the ordinate, , of the point to answer "What is the maximum..."

This might make more sense if you put your function into standard form:



so that it is clear that 1: is a function of and 2: the lead coefficient is -16 and the linear term coefficient is 96]

You might also want to note that for this model to be an accurate representation of a physical phenomenon, the object must be touching the ground when it was launched with a 96 feet per second upward initial velocity.


John

My calculator said it, I believe it, that settles it