Question

Using the equation representing the height of the firework (h = -16t2 + v0t + h0), algebraically determine the extreme value of f(t) by completing the square and finding the vertex. Interpret what the value represents in this situation.

Answer 1

-16t^2 + v0t + h0

=  -16(t^2 - 1/16 v0t)  + h0
 = -16 [(t - 1/32v0)^2 - (1/32 v0)^2 ) + h0
 = -16(t - 1/32v0)^2 + 1/64 (v0)^2 + h0   which is the vertex form 

The vertex  is at the point (1/32v0, 1/64 (v0)^2)

1/32v0  represents the time  when the firework is at its maximum height and 1/64 (v0)^2  is this maximum height

Answer 2

Answer with explanation:

The equation representing the height of the firework is

  $h=-16t^2+v_{0}t+h_{0}\\\\h=-16(t^2-\frac{v_{0}t}{16}-\frac{h_{0}}{16})\\\\h=-16[(t-\frac{v_{0}}{32})^2-(\frac{v_{0}}{32})^2-\frac{h_{0}}{16}]\\\\h-16[(\frac{v_{0}}{32})^2+\frac{h_{0}}{16}]=-16[(t-\frac{v_{0}}{32})]^2\\\\ \text{This is a function in t and h}\\\\ \text{Vertex}=(\frac{v_{0}}{32},16[(\frac{v_{0}}{32})^2+\frac{h_{0}}{16})])$

To determine the extreme of the function , we will differentiate it once

     $h=-16t^2+v_{0}t+h_{0}\\\\\frac{dh}{dt}=-32 t+v_{0}\\\\\text{Put},\frac{dh}{dt}=0\\\\\rightarrow -32 t+v_{0}=0\\\\\rightarrow t=\frac{v_{0}}{32}\\\\\text{Extreme value}=-16*[\frac{v_{0}}{32}]^2+v_{0}*\frac{v_{0}}{32}+h_{0}\\\\f(t)=-16*[\frac{v_{0}}{32}]^2+\frac{(v_{0})^2}{32}+h_{0}\\\\\frac{d^2h}{dt^2}=-32\\\\\text{Showing that function attains maximum at this point}$

The Extreme Value represents maximum height obtained by Firework.