Answer:
The ball will take 4.05 seconds to hit the ground.
Step-by-step explanation:
we have
[tex]h=-16t^{2}+64t+3[/tex]
This is a quadratic equation (vertical parabola) open down
The vertex is a maximum
we know that
The ball hit the ground when h=0
Solve the quadratic equation
For h=0
[tex]-16t^{2}+64t+3=0[/tex]
The formula to solve a quadratic equation of the form
[tex]at^{2} +bt+c=0[/tex] is equal to
[tex]t=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]-16t^{2}+64t+3=0[/tex]
so
[tex]a=-16\\b=64\\c=3[/tex]
substitute in the formula
[tex]t=\frac{-64(+/-)\sqrt{64^{2}-4(-16)(3)}} {2(-16)}[/tex]
[tex]t=\frac{-64(+/-)\sqrt{4,288}} {-32}[/tex]
[tex]t=\frac{64(-)\sqrt{4,288}} {32}=-0.05\ sec[/tex] ---> the time cannot be a negative number
[tex]t=\frac{64(+)\sqrt{4,288}} {32}=4.05\ sec[/tex]
therefore
The ball will take 4.05 seconds to hit the ground.