SOLUTION: A ball is thrown straight upward at an initial speed of 203 feet per second. From Physics we know that the ball will reach a height of h feet after t seconds where h and t are rel

Question 400014: A ball is thrown straight upward at an initial speed of 203 feet per second.
From Physics we know that the ball will reach a height of h feet after t seconds where h and t are related by the following formula.
h=−16t2+203t
When does the ball reach a height of 25 feet?
If the ball reaches 25 feet more than once, list both times in any order and separate them with commas.
If the ball never reaches 25 feet, enter NEVER.

Found 2 solutions by nerdybill, josmiceli:
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
A ball is thrown straight upward at an initial speed of 203 feet per second.
From Physics we know that the ball will reach a height of h feet after t seconds where h and t are related by the following formula.
h=−16t^2+203t
When does the ball reach a height of 25 feet?
set h to 25 and solve for t:
25=−16t^2+203t
0=−16t^2+203t-25
applying the quadratic formula we get:
t = {0.12, 12.56} 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=39609 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 0.124371891016755, 12.5631281089832. Here's your graph:

Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!

What is when ?

use quadratic formula:

sec
and, using the (-) root,

sec
The ball reaches 25' at t = .12437 sec and 12.563 sec
To check the answer:
The peak of the curve should be exactly 1/2 way between
these times, and max =
sec
So,
sec
OK

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