Answer:
[tex]t=4s[/tex]
Explanation:
Given a quadratic equation in its standard form:
[tex]ax^2+bx+c=0[/tex]
It's roots can be found using the quadratic formula, given by:
[tex]x=\frac{-b\pm \sqrt{b^2-4ac} }{2a}[/tex]
That's what we need to do in this problem, because we need to know for what values of t h(t)=0:
[tex]h(t)=-16t^2+64t=0[/tex]
In this case:
[tex]a=-16\\b=64\\c=0[/tex]
So:
[tex]t=\frac{-64\pm \sqrt{64^2-(4*(-16)*0} }{-16*2} = \frac{-64\pm \sqrt{64^2-0} }{-32}=\frac{-64\pm 64}{-32} \\\\Hence\\\\t=\frac{-64+64}{-32}=\frac{0}{-32} =0\\\\or\\\\t=\frac{-64-64}{-32} =\frac{-128}{-32} =4[/tex]
We are not interested in the first solution t=0 because it is the time before the golf ball is hit from the ground. The answer we look is the second solution t=4.
Therefore, it takes 4 seconds to return to the ground.
