Given:
The height of ball is represented by the equation is
[tex]h=-16t^2+6t+4[/tex]
To find:
The time taken by ball to hit the ground.
Solution:
First we need to find the zeroes of the given equation. So, for zeroes h=0.
[tex]-16t^2+6t+4=0[/tex]
Here, [tex]a=-16,b=6,c=4[/tex]. Usingquadratic formula, we get
[tex]t=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
[tex]t=\dfrac{-6\pm \sqrt{6^2-4(-16)(4)}}{2(-16)}[/tex]
[tex]t=\dfrac{-6\pm \sqrt{36+256}}{-32}[/tex]
[tex]t=\dfrac{-6\pm \sqrt{292}}{-32}[/tex]
[tex]t=-\dfrac{-6+ 17.088}{-32},\dfrac{-6+ 17.088}{-32}[/tex]
[tex]t=0.7215,-0.3465[/tex]
Time cannon be negative. So, at t=0.7215 the ball will hit the ground. Time taken by ball to hit the ground is
[tex]t=0.7215-0[/tex]
[tex]t=0.7215[/tex]
Round your answer to the nearest hundredth.
[tex]t\approx 0.72[/tex]
So, the ball will hit the ground after 0.72 seconds.
Therefore, the correct option is A.
