The path of a rocket follows the quadratic model: h = -162 +100t + 50, where h is
the height in feet given t seconds from launch. Find out how long, to the nearest
hundredth of a second, it takes the rocket to hit the ground. (if the rocket is on the
ground, it's height, h, is now zero, right ?! )

[tex]t = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\t = \frac{-(100)\pm\sqrt{(100)^2-4(-16)(50)}}{2(-16)}\\\\t = \frac{-100\pm\sqrt{13200}}{-32}\\\\t = \frac{-100+\sqrt{13200}}{-32} \ \text{ or } \ t = \frac{-100-\sqrt{13200}}{-32}\\\\t \approx -0.47 \ \text{ or } \ t \approx 6.72\\\\[/tex]

Ignore the negative solution because a negative time value does not make sense. The only reasonable solution is roughly t = 6.72

It takes approximately 6.72 seconds for the rocket to hit the ground.

The path of a rocket follows the quadratic model: h = -162 +100t + 50, where h is the height in feet given t seconds from launch. Find out how long, to the nearest hundredth of a second, it takes the rocket to hit the ground. (if the rocket is on the ground, it's height, h, is now zero, right ?! )

Respuesta :

Answer: 6.72 seconds

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The path of a rocket follows the quadratic model: h = -162 +100t + 50, where h is
the height in feet given t seconds from launch. Find out how long, to the nearest
hundredth of a second, it takes the rocket to hit the ground. (if the rocket is on the
ground, it's height, h, is now zero, right ?! )