Question

A small rock falling from the top of a 124-ft-tall building with an initial downward velocity of -30ft/sec is modeled by the equation h(t)= -16t2-30t+124, where t is the time in seconds. For which interval of time does the rock remain in the air?

Answer 1

You need to solve this for h(t)=0

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

This has two solutions, one is negative that does not make sense as time cannot be negative. 
The positive solution is t=2

So the interval where it is in the air is [0;2) 

Answer 2

Answer:

$0\leq t$

Step-by-step explanation:

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 .

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

Now we are supposed to find For which interval of time does the rock remain in the air

Substitute h(t)=0

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

$-2(8t^2+15t-62)=0$

$(t-2)(8t+31)=0$

$t = 2, \frac{-31}{8}$

Since time cannot be negative .So, neglect $\frac{-31}{8}$

So, time interval for which  the rock remain in the air:

$0\leq t$