Question 1158111: f(x) = −16x^2 + 96x
At what time does the ball hit the ground?
What is the maximum height of the ball?
After how many seconds does the ball reach its maximum height?
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
The proper way to express this function in general is:
\ =\frac{1}{2}gt^2\ +\ v_ot\ +\ h_o)
Where  , the vertical acceleration of an object near the earth's surface,  is the initial velocity, and  is the initial height.
In your case,  and 
The projectile is on the ground whenever the value of the function is zero, so \ =\ h(0)\ =\ 0) means the ball is on the ground at the time of launch, and \ =\ 0) means that the ball returns to the ground at time  .
\ =\ -16t^2\ +\ 96t)
Has two zeros, one at time zero or  . You can solve the quadratic equation
for the other.
The graph of the function is a parabola opening downward. The maximum height is the maximum value of the height function at the vertex of the parabola. The time in seconds for the ball to reach this height is the value of the independent variable at the vertex of the parabola. In other words, the vertex of the parabola is at the point )) where the ordinate of the point is the maximum height, and the abscissa is the time in seconds it takes to get there.
Hint: The vertex of a parabola of the form \ =\ ax^2\ +\ bx\ +\ c) is found at the point \))
That is all the information you need to solve this problem for yourself.
John
My calculator said it, I believe it, that settles it
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