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 $\frac{-b}{2a}$.
From our function we can infer that $b=80$ and $a=-16$, so the point \frac{-b}{2a} [/tex] will be $\frac{-80}{-2(16)} = \frac{5}{2}$. Lets evaluate the function at that point:
$h=-16t^2+80t+5$
$h=-16( \frac{5}{2} )^2+80( \frac{5}{2} )+5$
$h=105$

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:
$h=-16t^2+80t+5$
$105=-16t^{2}+80t+5$
$-16t^2+80t-100=0$
$-4(4t^{2}+20t-25)=0$
$4(2t-5)^2=0$
$2t-5=0$
$t= \frac{5}{2}$
$t=2.5$

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.