Respuesta :
-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
= -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 with explanation:
The equation representing the height of the firework is
[tex]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})])[/tex]
To determine the extreme of the function , we will differentiate it once
[tex]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}[/tex]
The Extreme Value represents maximum height obtained by Firework.
