Question

a ball is thrown from a height of 79 meters with an initial downward velocity of 6 m/s, the balls height ( h ) (in meters) after ( t ) seconds is given by the following.
( h = 79 - 6t - 5t^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.)

Explanation:

Step1: Set height to 0 (ground)

$0 = 79 - 6t - 5t^2$
Rearrange to standard quadratic form:
$5t^2 + 6t - 79 = 0$

Step2: Apply quadratic formula

For $at^2+bt+c=0$, $t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. Here $a=5$, $b=6$, $c=-79$.

$$ t=\frac{-6\pm\sqrt{6^2-4(5)(-79)}}{2(5)} $$

Step3: Calculate discriminant

$\sqrt{36 + 1580} = \sqrt{1616} \approx 40.20$

Step4: Solve for positive t

Only positive time is valid:
$t=\frac{-6 + 40.20}{10} = \frac{34.20}{10} = 3.42$

Answer:

3.42 seconds