Question

an object is propelled vertically at an initial speed of 160 feet per second. determine the time it will take for the object to reach a height of 200 feet. round to the nearest tenth of a second. select one: a. t ≈ 1 seconds, t ≈ 9 seconds b. t ≈ 2.5 seconds, t ≈ 7.5 seconds c. t ≈ 1.5 seconds, t ≈ 8.5 seconds d. t ≈ 1.9 seconds, t ≈ 8.1 seconds

Explanation:

Step1: Use vertical motion formula

The vertical motion formula for height is $h(t) = -16t^2 + v_0t + h_0$, where $v_0=160$ ft/s, $h_0=0$ ft, $h(t)=200$ ft. Substitute values:
$$200 = -16t^2 + 160t$$

Step2: Rearrange to standard quadratic form

Bring all terms to one side:
$$16t^2 - 160t + 200 = 0$$
Divide by 4 to simplify:
$$4t^2 - 40t + 50 = 0$$

Step3: Apply quadratic formula

Quadratic formula: $t = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$, where $a=4$, $b=-40$, $c=50$.
First calculate discriminant:
$$\Delta = (-40)^2 - 4(4)(50) = 1600 - 800 = 800$$
Then solve for $t$:
$$t = \frac{40 \pm \sqrt{800}}{8} = \frac{40 \pm 20\sqrt{2}}{8} = \frac{20 \pm 10\sqrt{2}}{4}$$

Step4: Compute numerical values

$\sqrt{2} \approx 1.414$, so:
$t_1 = \frac{20 + 10(1.414)}{4} = \frac{34.14}{4} \approx 8.1$
$t_2 = \frac{20 - 10(1.414)}{4} = \frac{5.86}{4} \approx 1.5$
(Note: Correction: $\frac{20-14.14}{4}=\frac{5.86}{4}\approx1.5$, $\frac{20+14.14}{4}=\frac{34.14}{4}\approx8.1$)

Answer:

C. t ≈ 1.5 seconds, t ≈ 8.5 seconds
(Note: The precise calculation gives ~1.5 and ~8.1, option C is the closest match among choices)