Open Question: ALGEBRA II: A ball is thrown from a building 40 feet above ground level with initial vertical speed of...?
The height the ball is above the ground is given by the equation:
h = -16x^2 + 12x + 40
h = -16x^2 + 12x + 40
This already factors in gravitational acceleration and the original velocity (I am in physics).
Therefore, you just solve for the time when h= 0 (the ground).
0 = -16x^2 + 12x + 40
Divide both sides by -4 (to simplify)
0 = 4x^2 - 3x - 10
Factor:
0 = (x-2)(4x+5)
If you solve for x, you get that:
x= -(4/5), or 2
Disregard the negative answer, and you get that x = 2 seconds.
The correct answer is B
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