Question

Practice: The crew from Extravaganza Entertainment launches fireworks at an angle of 60degrees. The height of one display can be approximated by: h(t) = -16t2 + 128V3t,where h(t) is measured in feet and t in seconds.

Answer 1

Solution

Given

$h(t)=-16t^2+128\sqrt[]{3}t$

Analyse

The function is quadratic and to get the maximum height as well as the time taken to get the maximum height, we will have to write h(t) in this form

$h(t)=a(t-b)^2+c$

Solve

$\begin{gathered} h(t)=-16t^2+128\sqrt[]{3}t \\ h(t)=-16(t^2-8\sqrt[]{3}t) \\ h(t)=-16\lbrack t^2-8\sqrt[]{3}t+(4\sqrt[]{3})^2-(4\sqrt[]{3})^2\rbrack \\ h(t)=-16\lbrack t^2-8\sqrt[]{3}t+(4\sqrt[]{3})^2\rbrack+16(4\sqrt[]{3})^2 \\ h(t)=-16(t-4\sqrt[]{3})^2+768 \end{gathered}$$\begin{gathered} time=4\sqrt[]{3}=6.9282\text{ secs} \\ time=7\text{ secs (to the nearest second)} \\ MaxHeight=\text{ 768 fe}ets \end{gathered}$

Paraphrase

Therefore, It would take the fireworks 7 seconds to get to it's maximum height and the Maximum Height of the is 768 feets