Question

You are jumping on a trampoline. For one jump, your height y (in feet) above the trampoline after t seconds can be represented by y= -16t^2+24t. How many seconds are you in the air?

Answer 1

`1.5\; \rm s`.

Explanation:

The person in this question is in the air whenever the height is greater than
`0`.

The graph of
`y = -16\, t^(2) + 24\, t` is a parabola opening downwards. Rearrange and find values of
`t` that would set this expression to
`0`.

`\begin{aligned}y &= -16\, t^2 + 24\, t \\ &= t\, (-16\, t + 24) = -8\, t\, (2\, t - 3)\end{aligned}`.

The first factor,
`t`, suggests that
`t = 0` would set this expression to
`0`- quite expected, since the person is on the ground right before jumping.

The second factor,
`(2\, t - 3)`, suggests that
`2\, t = 3` (in other words,
`t = 3/2 = 1.5`) would also set this expression to
`0`. Hence, this person would be once again on the ground
`1.5` seconds after jumping.

Hence, this person is in the air between
`t = 0` and
`t= 1.5` for a total of
`1.5\; \rm s`.