The time taken by football to hit the ground is:
t=2.3561 sec.
The height(h) of the football at a time t is given by the equation:
[tex]h=-16t^2+36t+4[/tex]
Now, we are asked to find the time the ball will take to hit the ground.
i.e. we are asked to find the value of t when h=0
i.e.
[tex]-16t^2+36t+4=0\\\\i.e.\\\\16t^2-36t-4=0[/tex]
( Since, on multiplying both the sides of the equation by -1 )
Now, we divide both the side of the equation by 4 to get:
[tex]4t^2-9t-1=0[/tex]
We know that the solution of the quadratic equation of the type:
[tex]ax^2+bx+c=0[/tex]
is given by the formula:
[tex]x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
Here we have:
[tex]a=4,\ b=-9\ and\ c=-1[/tex]
Hence, the solution is:
[tex]t=\dfrac{-(-9)\pm \sqrt{(-9)^2-4\times 4\times (-1)}}{2\times 4}\\\\i.e.\\\\t=\dfrac{9\pm \sqrt{81+16}}{8}\\\\i.e.\\\\t=\dfrac{9\pm \sqrt{97}}{8}\\\\i.e.\\\\t=\dfrac{9+\sqrt{97}}{8},\ t=\dfrac{9-\sqrt{97}}{8}[/tex]
Also, we know that:
[tex]\sqrt{97}=9.8489[/tex]
This means that:
[tex]t=\dfrac{18.8489}{8}\, t=\dfrac{-0.8489}{8}\\\\i.e.\\\\t=2.3561,\ t=-0.1061[/tex]
time can't be negative.
Hence, the solution is: [tex]t=2.3561\ sec.[/tex]
