SOLUTION: When a ball is thrown, its height in feet h after t seconds is given by the equation vt-16t^2 where v is the initial upwards velocity in feet per second. If v=35 feet per second, f

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Question 332089: When a ball is thrown, its height in feet h after t seconds is given by the equation vt-16t^2 where v is the initial upwards velocity in feet per second. If v=35 feet per second, find all values of t for which h=18 feet
Answer by nerdybill(7384) (Show Source):
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When a ball is thrown, its height in feet h after t seconds is given by the equation vt-16t^2 where v is the initial upwards velocity in feet per second. If v=35 feet per second, find all values of t for which h=18 feet
.
The problem gives us:
h = vt-16t^2
and v=35
h = 35t-16t^2
.
Now, when h=18
18 = 35t-16t^2
Solving for t:
18 = 35t-16t^2
16t^2+18 = 35t
16t^2-35t+18 = 0
Applying the "quadratic formula" we find that:
t = {0.83, 1.36} seconds
.
Details 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=73 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 1.36075011704117, 0.826749882958827. Here's your graph: