A ball is thrown into the air with an upward velocity of 80ft/s. Its height H in feet after T seconds id given by the function H= -16T^2+80T+5. What is the maximum height the ball reaches? How long will it take the ball to reach maximum height? How long does it take for the mall to be caught 5 feet off the ground?All I really need help on is the last one.Will some one plz help me.

1. The function H= -16T^2+80T+5 is a parabola of the form [tex]a x^{2} +bx+c[/tex], so to find the maximum height of the ball, we are going to find the y-coordinate of the vertex of the parabola. To find the y-coordinate of the vertex we are going to evaluate the function at the point [tex] \frac{-b}{2a} [/tex].
From our function we can infer that [tex]b=80[/tex] and [tex]a=-16[/tex], so the point \frac{-b}{2a} [/tex] will be [tex] \frac{-80}{-2(16)} = \frac{5}{2} [/tex]. Lets evaluate the function at that point:
[tex]h=-16t^2+80t+5[/tex]
[tex]h=-16( \frac{5}{2} )^2+80( \frac{5}{2} )+5[/tex]
[tex]h=105[/tex]

We can conclude that the ball reaches a maximum height of 105 feet.

2. Since we now know that the maximum height the ball reaches is 105 feet, we are going to replace [tex]h[/tex] with 105 in our function, then we are going to solve for [tex]t[/tex] to find how long the ball takes to reach its maximum height:
[tex]h=-16t^2+80t+5[/tex]
[tex]105=-16t^{2}+80t+5[/tex]
[tex]-16t^2+80t-100=0[/tex]
[tex]-4(4t^{2}+20t-25)=0[/tex]
[tex]4(2t-5)^2=0[/tex]
[tex]2t-5=0[/tex]
[tex]t= \frac{5}{2} [/tex]
[tex]t=2.5[/tex]

We can conclude that the ball reaches its maximum height in 2.5 seconds.

3. Just like before, we are going to replace [tex]h[/tex] with 5 in our original function, then we are going to solve for [tex]t[/tex] to find how long will take for the ball to be caught 5 feet off the ground:
[tex]h=-16t^2+80t+5[/tex]
[tex]5=-16t^2+80t+5[/tex]
[tex]-16t^{2}+80t=0[/tex]
[tex]-16t(t-5)=0[/tex]
[tex]t-5=0[/tex]
[tex]t=5[/tex]

We can conclude that it takes 5 seconds for the ball to be caught 5 feet off the ground.