Question

Kehlani launches a toy rocket from a platform. The height of the rocket in feet is given by h= -16t^2 + 72t + 144 where t represents the time in seconds after launch. How long is the rocket in the air?

Answer 1

6 seconds

Explanation:

Given function:

`h(t)=-16t^2+72t+144`

where:

• h is the height of the rocket.
• t is the time (in seconds) after launch.

To find how long the rocket is in the air, find the positive value of t when h(t)=0.

`\implies -16t^2+72t+144=0`

`\implies -8(2t^2-9t-18)=0`

`\implies 2t^2-9t-18=0`

`\implies 2t^2+3t-12t-18=0`

`\implies t(2t+3)-6(2t+3)=0`

`\implies (t-6)(2t+3)=0`

Apply the zero-product property:

`t-6=0 \implies t=6`

`2t+3=0 \implies t=-(3)/(2)`

Therefore, as t > 0, the rocket is in the air for 6 seconds.

Answer 2

• 6 seconds

--------------------------------

Rocket is in the air until it hits the ground.

Therefore we need to find the value of t when h = 0.

• - 16t² + 72t + 144 = 0
• 2t² - 9t - 18 = 0
• 2t² - 12t + 3t - 18 = 0
• 2t(t - 6) + 3(t - 6) = 0
• (t - 6)(2t + 3) = 0
• t - 6 = 0 or 2t + 3 = 0
• t = 6 or t = - 3/2 (discarded as negative)

The rocket is in the air for 6 seconds.