Answer:
1.74 second long after it was thrown does it hit the ground.
Step-by-step explanation:
Given : A football is thrown with an initial upward velocity of 25 feet per second from a height of 5 feet above the ground. The equation [tex]h =-16t^2+25t+5[/tex] models the height in feet t seconds after it is thrown.
To find : How long after it was thrown does it hit the ground?
Solution :
The equation model is [tex]h(t)=-16t^2+25t+5[/tex]
After the ball passes its maximum height, it comes down and hits the ground.
i.e. h=0
So, [tex]-16t^2+25t+5=0[/tex]
Solve by quadratic formula of equation [tex]ax^2+bx+c=0[/tex] is [tex]x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]
Here, a=-16 , b=25 and c=5
[tex]t=\frac{-25 \pm \sqrt{(25)^2-4(-16)(5)}}{2(-16)}[/tex]
[tex]t=\frac{-25 \pm \sqrt{625+320}}{-32}[/tex]
[tex]t=\frac{-25 \pm \sqrt{945}}{-32}[/tex]
[tex]t=\frac{-25+\sqrt{945}}{-32},\frac{-25-\sqrt{945}}{-32}[/tex]
[tex]t=-0.179,1.741[/tex]
We reject t=-0.179.
Therefore, 1.74 second long after it was thrown does it hit the ground.
