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) = -16t² + 40t + 1.5
your answer
this is a required question

Explanation:

Step1: Set height equal to 20

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

Step2: Rearrange to standard quadratic form

$-16t^2 + 40t + 1.5 - 20 = 0$
$-16t^2 + 40t - 18.5 = 0$
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 $a=16$, $b=-40$, $c=18.5$
First calculate discriminant:
$\Delta = (-40)^2 - 4(16)(18.5) = 1600 - 1184 = 416$
$\sqrt{416} = \sqrt{16\times26} = 4\sqrt{26} \approx 20.396$

Step4: Calculate two solutions

$t_1 = \frac{40 + 20.396}{2(16)} = \frac{60.396}{32} \approx 1.89$
$t_2 = \frac{40 - 20.396}{2(16)} = \frac{19.604}{32} \approx 0.61$

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).