A model rocket is launched with an initial upward velocity of 151 ft/s. The rockets height "h" (in feet) after "t" seconds is given by the following. H=151t-16t squared.Find the values of "t" for which the rockets height is 88 feet.Round your answer to the nearest hundredth.
Accepted Solution
A:
Answer:t=8.81s and t=0.62sStep-by-step explanation:Asking for H=88ft and putting units leaves us with the formula:[tex]88ft=(151ft/s)t-(16ft/s^2)t^2[/tex]Which can be also written as:[tex](16ft/s^2)t^2-(151ft/s)t+88ft=0[/tex]We want t, and this is a quadratic formula of the form [tex]at^2+bt+c=0[/tex], which we know has the solutions:[tex]t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]So we put the relevant values (a=16ft/s^2, b=-151ft/s, c=88ft) on that equation we get:[tex]t=\frac{(151ft/s)\pm\sqrt{(-151ft/s)^2-4(14ft/s^2)(88ft)}}{2(16ft/s2)}[/tex]Which for the plus sign gives t=8.81345408729s and for the minus sign gives t=0.6240459127s, which rounding to the nearest hundredth are t=8.81s and t=0.62s