SOLUTION: An object is launched into the air from a ledge 6 feet off the ground at initial vertical velocity of 96 feet per second. Its height H, in feet, at t second is given by the equatio

Question 1087634: An object is launched into the air from a ledge 6 feet off the ground at initial vertical velocity of 96 feet per second. Its height H, in feet, at t second is given by the equation H = 16t2+96t+16. Find all the times t that the object is at a height of 160 feet off the ground.
Found 2 solutions by ikleyn, Alan3354:
Answer by ikleyn(52781) (Show Source): You can put this solution on YOUR website!
.
I see at least TWO errors in your formula.

So, carefully check your post.

Read and re-read it until you find the errors.

Then fix them.

Then re-post.

Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
An object is launched into the air from a ledge 6 feet off the ground at initial vertical velocity of 96 feet per second. Its height H, in feet, at t second is given by the equation H = 16t2+96t+16. Find all the times t that the object is at a height of 160 feet off the ground.
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H = 16t2+96t+16
H = 16t2+96t+16 = 96
16t2+96t+16 = 96
16t2+96t-80 = 0
t^2 + 6t - 80 = 0
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Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=356 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 6.4339811320566, -12.4339811320566. Here's your graph:

===========
t = ~6.433 seconds (based on your info)
=============================
You might notice that the object's height increases with time. It never "comes down" or impacts.
That sorta makes sense if it'a rocket with thrust, but I doubt that's what the problem stated.

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