Answer:
3 seconds
Step-by-step explanation:
The height of the rocket when it hits the ground will be zero meters.
Therefore, set the equation to zero and solve for t by using the quadratic formula.
Quadratic Formula
[tex]x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0[/tex]
Given equation:
[tex]-4.9t^2+14t+2.1=0[/tex]
Therefore:
[tex]a=-4.9, \quad b=14, \quad c=2.1[/tex]
Substitute the values into the quadratic formula:
[tex]\implies t=\dfrac{-14 \pm \sqrt{14^2-4(-4.9)(2.1)}}{2(-4.9)}[/tex]
[tex]\implies t=\dfrac{-14 \pm \sqrt{196+41.16}}{-9.8}[/tex]
[tex]\implies t=\dfrac{-14 \pm \sqrt{237.16}}{-9.8}[/tex]
[tex]\implies t=\dfrac{-14 \pm15.4}{-9.8}[/tex]
Therefore:
[tex]\implies t=\dfrac{-14 +15.4}{-9.8}=\dfrac{1.4}{-9.8}=-\dfrac{1}{7}[/tex]
[tex]\implies t=\dfrac{-14-15.4}{-9.8}=\dfrac{-29.4}{-9.8}=3[/tex]
As time is positive, t = 3 s only.
Therefore, it will take 3 seconds for the rocket to hit the ground.
