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

`0\leq t <2`

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, (-31)/(8)`

Since time cannot be negative .So, neglect
`(-31)/(8)`

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

`0\leq t <2`