Question

the height, h (in feet) of a model rocket launched from the roof of a building at t seconds is given by h = s(t)= - 16(t + 2)(t - 6)
a. expand s(t)= - 16(t + 2)(t - 6) to write it in the form s(t)=at^2 + bt + c.
s(t)= - 16t^2+64t + 192
b. what is the height of the rocket at t = 0?
the inital height is feet

Explanation:

Step1: Expand the expression

First, expand \((t + 2)(t - 6)\) using FOIL method: \((t+2)(t - 6)=t^{2}-6t + 2t-12=t^{2}-4t - 12\). Then multiply by - 16: \(s(t)=-16(t^{2}-4t - 12)=-16t^{2}+64t + 192\).

Step2: Find the initial height

Substitute \(t = 0\) into \(s(t)=-16t^{2}+64t + 192\). We get \(s(0)=-16\times0^{2}+64\times0 + 192=192\).

Answer:

b. 192