SOLUTION: If an object is given an initial velocity of v0 feet per second from a height of s0 feet, then its height as a function of time is h(t) = -16t^2 + v0t + s0 A potato gun is ca

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: If an object is given an initial velocity of v0 feet per second from a height of s0 feet, then its height as a function of time is h(t) = -16t^2 + v0t + s0 A potato gun is ca      Log On


   

Question 194477: If an object is given an initial velocity of v0 feet per second from a height of s0 feet, then its height as a function of time is
h(t) = -16t^2 + v0t + s0
A potato gun is capable of launching a spud with an initial velocity of 80 feet per second. If the tater takes off from an elevation of 100 feet, how many seconds into its flight will it be at an elevation of 115 feet.
Found 2 solutions by Alan3354, nerdybill:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
h(t) = -16t^2 + v0t + s0
A potato gun is capable of launching a spud with an initial velocity of 80 feet per second. If the tater takes off from an elevation of 100 feet, how many seconds into its flight will it be at an elevation of 115 feet.
-----------------------------
h(t) = -16t^2 + 80t + 100
At 115 feet:
115 = -16t^2 + 80t + 100
16t^2 - 80t + 15 = 0
t = (10 ± sqrt(85))/4 seconds
One solution for t is on its way up, the other is on its way down.
t = ~ 0.19511 seconds going up
t = ~ 4.80489 seconds going down

Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
Simply plug in the given information into:
h(t) = -16t^2 + v0t + s0
.
h(t) = -16t^2 + v0t + s0
115 = -16t^2 + 80t + 100
Moving all terms to the left:
16t^2 - 80t + 15 = 0
Solving the above using the quadratic equation yields:
x = {4.805, 0.195}
.
The tater is at 115 ft at 0.195 secs and again at 4.805 secs after flight.
.
Details of quadratic follows:
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 16x%5E2%2B-80x%2B15+=+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=%28-80%29%5E2-4%2A16%2A15=5440.

Discriminant d=5440 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--80%2B-sqrt%28+5440+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-80%29%2Bsqrt%28+5440+%29%29%2F2%5C16+=+4.80488611432322
x%5B2%5D+=+%28-%28-80%29-sqrt%28+5440+%29%29%2F2%5C16+=+0.195113885676778

Quadratic expression 16x%5E2%2B-80x%2B15 can be factored:
16x%5E2%2B-80x%2B15+=+16%28x-4.80488611432322%29%2A%28x-0.195113885676778%29
Again, the answer is: 4.80488611432322, 0.195113885676778. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+16%2Ax%5E2%2B-80%2Ax%2B15+%29