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) = -16% - 30t + 124, where t is the time in seconds. For which interval of time does the rock remain in the air?

O

t = 2

-2
0
t> 2

Answer 1

Answer:

For the time interval 0 to 2 the ball is in the air.

Select the options according to this.

Step-by-step explanation:

The equation should be $h(t)=-16t^{2} -30t+124$

Let's find the x intercepts to answer this question.

To find x intercept where the h(t) is 0.

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

Solve the equation for t.

Use quadratic formula to solve the equation.

a=-16

b=-30

c=124

$x=\frac{-b+/- \sqrt{b^{2}-4ac } }{2a}$

Plug in the values of a, b and c into the formula

$x=\frac{30+/-\sqrt{(-30)^{2}-4(-16)(124) } }{2(-16)}$

Simplify it

$x=\frac{30+/- \sqrt{900+7936} }{-32}$

$x=\frac{30+/-\sqrt{8836} }{-32}$

$x=\frac{30+/-94}{-32}$

Simplify it to get two values for x

$x=\frac{30+94}{-32} =-3.875\\\\x=\frac{30-94}{-32} =2$

x can not be negative, here x represents time t.

So, it starts from 0 and remains in the air for 2 seconds.