Question

A model rocket is launched with an initial upward velocity of 202 ft/s. The rocket’s height h (in feet) after t seconds is given by the following.h=202t-16t^2Find the value of t for which the rocket’s height is 82 feet. Round your answers to the nearest hundredth

Answer 1

Solution:

Given:

$h=202t-16t^2$$\begin{gathered} when\text{ the height is 82 feet,} \\ h=82 \\ \\ Hence, \\ h=202t-16t^2 \\ 82=202t-16t^2 \\ Collecting\text{ all terms to one side to make it a quadratic equation;} \\ 16t^2-202t+82=0 \\ Dividing\text{ the equation all through by 2,} \\ 8t^2-101t+41=0 \\ \\ To\text{ solve for t, we use the quadratic formula;} \\ \frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ where; \\ a=8,b=-101,c=41 \\ Hence, \\ t=\frac{-\left(-101\right)\pm\sqrt{\left(-101\right)^2-\left(4\times8\times41\right)}}{2\times8} \\ t=\frac{101\pm\sqrt{10201-1312}}{16} \\ t=\frac{101\pm\sqrt{8889}}{16} \\ t=\frac{101\pm94.28}{16} \\ t_1=\frac{101+94.28}{16}=\frac{195.28}{16}=12.205\approx12.21s \\ t_2=\frac{101-94.28}{16}=\frac{6.72}{16}=0.42s \end{gathered}$

Therefore, to the nearest hundredth, the value of t for which the rocket's height is 82 feet is;

0.42 seconds or 12.21 seconds.