Answer:
3.4 seconds
Step-by-step explanation:
Given function:
[tex]h(t)=-4.9t^2+2.45t+48.314[/tex]
where
The ball will strike the ground when its height is zero.
Therefore, to calculate when the ball strikes the ground, substitute h(t) = 0 and solve for t using the quadratic formula.
[tex]\boxed{\begin{minipage}{3.6 cm}\underline{Quadratic Formula}\\\\$x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$\\\\when $ax^2+bx+c=0$ \\\end{minipage}}[/tex]
Therefore:
Substitute these values into the quadratic formula:
[tex]\implies t=\dfrac{-2.45 \pm \sqrt{(2.45)^2-4(-4.9)(48.314)}}{2(-4.9)}[/tex]
[tex]\implies t=\dfrac{-2.45 \pm \sqrt{6.0025+946.9544}}{-9.8}[/tex]
[tex]\implies t=\dfrac{-2.45 \pm \sqrt{952.9569}}{-9.8}[/tex]
[tex]\implies t=\dfrac{-2.45 \pm 30.87}{-9.8}[/tex]
[tex]\implies t=\dfrac{-2.45 + 30.87}{-9.8}=-2.9[/tex]
[tex]\implies t=\dfrac{-2.45 -30.87}{-9.8}=3.4[/tex]
As time cannot be negative, t = 3.4 s only.
Therefore, the ball strikes the ground 3.4 seconds after it is launched.