SOLUTION: Original equation= h=-5t^2+25t+80 The only part of the height equation that we have not discussed is the constant. You have probably noticed that the constant is always equal t

Question 39287This question is from textbook beggining algebra
: Original equation= h=-5t^2+25t+80
The only part of the height equation that we have not discussed is the constant. You have
probably noticed that the constant is always equal to the initial height of the ball (80 m in
our previous exercises). Now, let�s have you develop a height equation.
A ball is thrown upward from the roof of a building 100 m tall with an initial velocity
of 20 m/s. Use this information for exercises 35 to 38.
36. Science and medicine. When will the ball reach a height of 80 m? This question is from textbook beggining algebra

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
h=-5t^2+25t+100
Let h=80 and solve for "t" as follows:
80=-5t^2+25t+100
-5t^2+25t+20=0

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=1025 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: -0.701562118716424, 5.70156211871642. Here's your graph:

t=5.7... seconds is the only practical answer.
Cheers,
Stan H.
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