Question

a ball is thrown from a height of 205 feet with an initial downward velocity of 11 ft/s. the balls height h (in feet) after t seconds is given by the following.
\\( h = 205 - 11t - 16t^2 \\)
how long after the ball is thrown does it hit the ground?
round your answer(s) to the nearest hundredth.
if there is more than one answer, use the \or\ button.
image of a building, a ball, and a height label h to the ground

Explanation:

Step1: Set height to 0 (ground level)

$0 = 205 - 11t - 16t^2$
Rearrange to standard quadratic form:
$16t^2 + 11t - 205 = 0$

Step2: Apply quadratic formula

For $at^2+bt+c=0$, $t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. Here $a=16$, $b=11$, $c=-205$.
$$t=\frac{-11\pm\sqrt{11^2-4(16)(-205)}}{2(16)}$$

Step3: Calculate discriminant

$\sqrt{121 + 13120} = \sqrt{13241} \approx 115.07$

Step4: Solve for positive t

Only positive time is valid:
$t=\frac{-11 + 115.07}{32} \approx \frac{104.07}{32} \approx 3.25$

Answer:

3.25 seconds