Answer :
Answer: 1.4 seconds
Step-by-step explanation:
The equation is: h(t) = at² + v₀t + h₀ where
- a is the acceleration (in this case it is gravity)
- v₀ is the initial velocity
- h₀ is the initial height
Given:
- a = -9.81 (if it wasn't given in your textbook, you can look it up)
- v₀ = 12
- h₀ = 3
Since we are trying to find out when it lands on the ground, h(t) = 0
EQUATION: 0 = 9.81t² + 12t + 3
Use the quadratic equation to find the x-intercepts
a=-9.81, b=12, c=3
[tex]x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\\\\\\x=\dfrac{-(12)\pm \sqrt{(12)^2-4(-9.81)(3)}}{2(-9.81)}\\\\\\x=\dfrac{-12\pm 16.2}{-19.62}\\\\\\x=\dfrac{-12+ 16.2}{-19.62}=-0.2\qquad x=\dfrac{-12- 16.2}{-19.62}=\large\boxed{1.4}\\[/tex]
Note: Negative time (-0.2) is not valid