Question

Ramona launches a model rocket and watches as it goes up and then back to the ground. The height of the rocket after it is launched can be modeled by the quadratic function h = –16t2 + 96t + 3 , where h is the height in feet, and t is the number of seconds after Ramona launches the rocket. How long will the rocket be in the air before it hits the ground? Round to the nearest hundredth.

Answer 1

We want to see how much time will pass until the rocket falls to the ground again.

The answer is 6.03 seconds.

We know that the height equation of the rocket is:

h(t) = -16*t^2 + 96*t + 3

The rocket will hit the ground when it's height is equal to zero, then we must solve:

h(t) = 0 = -16*t^2 + 96*t + 3

So we just need to solve a simple quadratic equation, the solutions are given by Bhaskara's formula:

$t = \frac{-96 \pm \sqrt{96^2 - 4*3*(-16)} }{2*-16} \\\\t = \frac{-96 \pm 96.99 }{-32}$

Because t represents the time after the rocket is launched, we only care for positive values of t, then we only care for the positive solution, which is:

$t = \frac{-96 - 96.99 }{-32} = 6.03$

This means that after 6.03 seconds the rocket will hit the ground, so, the rocket will be on the air for 6.03 seconds.

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