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

Answer:

$1.5\; \rm s$.

Step-by-step 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 $1.5\; \rm s$.