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

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;} \\ (-b\pm√(b^2-4ac))/(2a) \\ where; \\ a=8,b=-101,c=41 \\ Hence, \\ t=(-\left(-101\right)\pm√(\left(-101\right)^2-\left(4*8*41\right)))/(2*8) \\ t=(101\pm√(10201-1312))/(16) \\ t=(101\pm√(8889))/(16) \\ t=(101\pm94.28)/(16) \\ t_1=(101+94.28)/(16)=(195.28)/(16)=12.205\approx12.21s \\ t_2=(101-94.28)/(16)=(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.