Question

a ball is thrown from a height of 151 feet with an initial downward velocity of 15 ft/s. the balls height h (in feet) after t seconds is given by the following. h = 151 - 15t - 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 to 0 (ground level)

$0 = 151 - 15t - 16t^2$
Rearrange to standard quadratic form:
$16t^2 + 15t - 151 = 0$

Step2: Apply quadratic formula

For $at^2+bt+c=0$, $t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.
Here $a=16$, $b=15$, $c=-151$.

$$ t=\frac{-15\pm\sqrt{15^2-4(16)(-151)}}{2(16)} $$

Step3: Calculate discriminant

$\sqrt{225 + 9664} = \sqrt{9889} \approx 99.4435$

Step4: Solve for positive t

We discard the negative root (time can't be negative):

$$ t=\frac{-15 + 99.4435}{32} \approx \frac{84.4435}{32} \approx 2.64 $$

Answer:

$t=2.64$ seconds