Question

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.

Answer 1

1. The function H= -16T^2+80T+5 is a parabola of the form
`a x^(2) +bx+c`, 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
`(-b)/(2a)`.
From our function we can infer that
`b=80` and
`a=-16`, so the point \frac{-b}{2a} [/tex] will be
`(-80)/(-2(16)) = (5)/(2)`. Lets evaluate the function at that point:

`h=-16t^2+80t+5`

`h=-16( (5)/(2) )^2+80( (5)/(2) )+5`

`h=105`

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
`h` with 105 in our function, then we are going to solve for
`t` to find how long the ball takes to reach its maximum height:

`h=-16t^2+80t+5`

`105=-16t^(2)+80t+5`

`-16t^2+80t-100=0`

`-4(4t^(2)+20t-25)=0`

`4(2t-5)^2=0`

`2t-5=0`

`t= (5)/(2)`

`t=2.5`

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

3. Just like before, we are going to replace
`h` with 5 in our original function, then we are going to solve for
`t` to find how long will take for the ball to be caught 5 feet off the ground:

`h=-16t^2+80t+5`

`5=-16t^2+80t+5`

`-16t^(2)+80t=0`

`-16t(t-5)=0`

`t-5=0`

`t=5`

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