Question

The height of a small rock falling from the top of a 124-ft-tall building with an initial downward velocity of –30 ft/sec is modeled by the equation h(t) = –16t^2 – 30t + 124, where t is the time in seconds. For which interval of time does the rock remain in the air?
t = 2

–2 < t < 0

0 < t < 2

t > 2

Answer 1

Answer:

Option 3.

Step-by-step explanation:

The given function is

$h(t)=-16t^2-30t+124$

where, h(t) is the height of a small rock falling from the top of a 124-ft-tall building and t is the time in seconds.

It is a downward parabola.

Equate the function equal to 0, to find the time at which the rock touch the ground.

$-16t^2-30t+124=0$

If $ax^2+bx+c=0$, the according to the quadratic formula

$x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

Using quadratic formula, we get

$t=\dfrac{-(-30)\pm \sqrt{(-30)^2-4(124)(-16)}}{2(-16)}$

$t=\dfrac{30+\sqrt{(-30)^2-4(124)(-16)}}{2(-16)},\dfrac{30-\sqrt{(-30)^2-4(124)(-16)}}{2(-16)}$

$t=-3.875, 2$

Time cannot be negative. So, the rock remain in the air in the interval 0 < t < 2.

Therefore, the correct option is 3.