SOLUTION: A ball is tossed into the air at an initial velocity of 38.4ft/sec from .96 feet off of the ground. At what point does the ball hit the ground? I figured out that the equatio

Question 390072: A ball is tossed into the air at an initial velocity of 38.4ft/sec from .96 feet off of the ground.
At what point does the ball hit the ground?
I figured out that the equation is y=-16t^2+38.4t+.96
I also figured out that the maximum height of the ball is 24 feet and it reaches it's maximum in 1.2 seconds.
Found 2 solutions by stanbon, nerdybill:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
You are correct.
Cheers,
Stan H.
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
y=-16t^2+38.4t+.96
set y to 0 and solve for t:
0=-16t^2+38.4t+.96
applying the quadratic equation:
t = {-0.02, 2.42}
throw out the negative solution, leaving:
t = 2.42 seconds
.

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=1536 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: -0.024744871391589, 2.42474487139159. Here's your graph:

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