Question

A model rocket is launched directly upward at a speed of 28 meters per second from a height of 12 meters. The function f(t) = - 4.9t ^ 2 + 28t + 12, mo models the relationship between the height of the rocket and the time after launch. in seconds When seconds after launch will the rocket hit the ground? Do not round until the final answer. Then round your answer to two decimal places. Provide your answer below:

Answer 1

Answer:

6.11 seconds

Step-by-step explanation:

Given

$f(t) = -4.9t^2 + 28t + 12$

Required

When will it hit the ground?

When it hits the ground, f(t) = 0.

So, we have:

$0 = -4.9t^2 + 28t + 12$

Solve quadratic equation using formula

$t = \frac{-b \± \sqrt{b^2 - 4ac}}{2a}$

$t = \frac{-28 \± \sqrt{28^2 - 4*-4.9*12}}{2*-4.9}$

$t = \frac{-28 \± \sqrt{1019.2}}{-9.8}$

$t = \frac{-28 \± 31.92}{-9.8}$

Split

$t = \frac{-28 + 31.92}{-9.8}$ or $t = \frac{-28 - 31.92}{-9.8}$

$t = \frac{3.92}{-9.8}$ or $t = \frac{-59.92}{-9.8}$

$t = -0.4$ or $t = 6.11$

Time can not be negative. So:

$t = 6.11$