Question

A rocket is launched and the quadratic function h(t) = -5t^2+ 165t relates the height (h) in meters to seconds (t) after launch. When will the rocket be 450 feet above the ground? (Hint...twice!!!)

Answer 1

Given:

$h(t)=-5t^2+165t$

where h is the height in meters and t is the second in seconds.

Required:

We need to find the time when the rocket is 450 feet height from the ground.

Explanation:

We need to convert 450 feet into meters.

$\text{We know that 1 foot = 0.3048 meters.}$

Multiply both sides by 450.

$450\text{ }\times\text{1 foot = 450 }\times\text{ 0.3048 meters.}$

$450\text{ }\times fee\text{t = 137.16 meters.}$

Substitute h(t)= 137.16 meters in the given equation to find the value of t.

$137.16=-5t^2+165t$

Solve for t.

$5t^2-165t+137.16=0$

Divide both sides of the equation by 5.

$\frac{5t^2}{5}-\frac{165t}{5}+\frac{137.16}{5}=0$

$t^2-33t+27.432=0$

which is of the form.

$at^2+bt+c=0$

where a =1, b=-33 and c =27.432.

Consider the quadratic formula.

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

Substitute a =1, b=-33, and c =27.432 in the formula to find the value of t.

$t=\frac{-(-33)\pm\sqrt{(-33)^2-4(1)(27.432)}}{2(1)}$

$t=\frac{33\pm\sqrt{1089-109.728}}{2}$

$t=\frac{33\pm\sqrt{979.272}}{2}$

$t=\frac{33\pm\sqrt{979.272}}{2}$