Question

a ball is thrown from a height of 125 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 = 125 - 11t - 16t²
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.)

Explanation:

Step1: Set height h to 0

$0 = 125-11t - 16t^{2}$

Step2: Rewrite in standard quadratic form

$16t^{2}+11t - 125=0$

Step3: Use quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

Here $a = 16$, $b = 11$, $c=- 125$. First calculate the discriminant $\Delta=b^{2}-4ac=(11)^{2}-4\times16\times(-125)=121 + 8000=8121$.
Then $t=\frac{-11\pm\sqrt{8121}}{2\times16}=\frac{-11\pm90.1166}{32}$.

Step4: Find two possible values of t

$t_1=\frac{-11 + 90.1166}{32}=\frac{79.1166}{32}\approx2.47$
$t_2=\frac{-11-90.1166}{32}=\frac{-101.1166}{32}\approx - 3.16$
Since time cannot be negative, we discard the negative - valued solution.

Answer:

$2.47$