Question

a ball was shot from a cannon into the air with an upward velocity of 40 ft/sec. the equation that gives the height (h) of the ball at any time (t) is shown below. how long does it take for the ball to reach a height of 20 feet?
( h(t) = -16t^2 + 40t + 1.5 )
your answer
this is a required question

Explanation:

Step1: Set h(t) equal to 20

$20 = -16t^2 + 40t + 1.5$

Step2: Rearrange into standard quadratic form

Subtract 20 from both sides:
$0 = -16t^2 + 40t - 18.5$
Multiply by -1:
$16t^2 - 40t + 18.5 = 0$

Step3: Apply quadratic formula

For $at^2+bt+c=0$, $t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. Here $a=16$, $b=-40$, $c=18.5$:
$$t=\frac{40\pm\sqrt{(-40)^2-4(16)(18.5)}}{2(16)}$$

Step4: Calculate discriminant

$\sqrt{1600 - 1184} = \sqrt{416} \approx 20.396$

Step5: Solve for t values

First solution:
$t=\frac{40 + 20.396}{32} \approx \frac{60.396}{32} \approx 1.89$
Second solution:
$t=\frac{40 - 20.396}{32} \approx \frac{19.604}{32} \approx 0.61$

Answer:

The ball reaches 20 feet at approximately $\boldsymbol{0.61}$ seconds (on the way up) and $\boldsymbol{1.89}$ seconds (on the way down).