SOLUTION: When a ball is thrown, its height in feet h after t seconds is given by the equation h=vt-16t^2 where v is the initial upwards velocity in feet per second. If v=9 feet per

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Question 350709: When a ball is thrown, its height in feet h after t seconds is given by the equation
h=vt-16t^2
where v is the initial upwards velocity in feet per second. If v=9 feet per second, find all values of t for which h=1 foot. Do not round any intermediate steps. Round your answer to 2 decimal places.
Answer by nerdybill(7384) (Show Source):
You can put this solution on YOUR website!
When a ball is thrown, its height in feet h after t seconds is given by the equation
h=vt-16t^2
where v is the initial upwards velocity in feet per second. If v=9 feet per second, find all values of t for which h=1 foot. Do not round any intermediate steps. Round your answer to 2 decimal places.
.
h=vt-16t^2
Substitute in the value for v:
h=9t-16t^2
Substitute in the value for h:
1=9t-16t^2
Solving for t:
1=9t-16t^2
16t^2+1=9t
16t^2-9t+1=0
Solve using the quadratic formula. Doing so yields
t = {0.41, 0.15} seconds
.
Detail of quadratic follows:

Solved by pluggable solver: SOLVE quadratic equation with variable

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=17 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 0.410097050800552, 0.152402949199448. Here's your graph: