Question

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.

Answer 1

Answer:

t=8.81s and t=0.62s

Step-by-step explanation:

Asking for H=88ft and putting units leaves us with the formula:

$88ft=(151ft/s)t-(16ft/s^2)t^2$

Which can be also written as:

$(16ft/s^2)t^2-(151ft/s)t+88ft=0$

We want t, and this is a quadratic formula of the form $at^2+bt+c=0$, which we know has the solutions:

$t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

So we put the relevant values (a=16ft/s^2, b=-151ft/s, c=88ft) on that equation we get:

$t=\frac{(151ft/s)\pm\sqrt{(-151ft/s)^2-4(14ft/s^2)(88ft)}}{2(16ft/s2)}$

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