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

Algebra ->  Quadratic Equations and Parabolas -> 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      Log On


   

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(52777) About Me  (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) About Me  (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.
---------------
H = 16t2+96t+16
H = 16t2+96t+16 = 96
16t2+96t+16 = 96
16t2+96t-80 = 0
t^2 + 6t - 80 = 0
-------
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B6x%2B-80+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%286%29%5E2-4%2A1%2A-80=356.

Discriminant d=356 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-6%2B-sqrt%28+356+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%286%29%2Bsqrt%28+356+%29%29%2F2%5C1+=+6.4339811320566
x%5B2%5D+=+%28-%286%29-sqrt%28+356+%29%29%2F2%5C1+=+-12.4339811320566

Quadratic expression 1x%5E2%2B6x%2B-80 can be factored:
1x%5E2%2B6x%2B-80+=+%28x-6.4339811320566%29%2A%28x--12.4339811320566%29
Again, the answer is: 6.4339811320566, -12.4339811320566. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B6%2Ax%2B-80+%29

===========
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.