The equation \(y=x^2-8x-33\) describes the height (in feet) of a ball tossed up in the air at 26 feet per second from a height of 105 feet above the ground. In how many seconds will the ball hit the ground? Express your answer as a decimal rounded to the nearest tenth.

imdumb Jul 13, 2020

#1**0 **

When the ball hits the ground, y will be = zero

zero 0 = x^2-8x-33 solve for x .... throw out any negative values

ElectricPavlov Jul 13, 2020

#6**+2 **

The solution for the equation may be correct, but…not *correct-o-mundo*.

**This equation is inconsistent with the question. **

Initially, the equation describes a ball thrown **downward** **at 8 feet per second**, from a height **33 feet below ground level**, in a **negative gravity** environment with an upward acceleration of -0.06216G –this corresponds to an acceleration of just under -2fts^{-2}.

After the ball slows to a stop, and rises, the velocity curve implies the gravity changes to -0.03108G –this corresponds to an acceleration of

just under -1fts^{-2 }

If the ball bounces in a perfectly elastic collision and rises at an initial velocity of 8 feet per second, then slowing to 7 feet per second as passes ground level, implies the acceleration from gravity is now positive @ 0.00444G (This implies gravity changed from negative to positive.)

Note that negative gravity implies negative mass. https://en.wikipedia.org/wiki/Negative_mass

**One possible environment where this could occur in the real world is in a liquid environment where the ball is positively buoyant. The change in pressure as the ball changes in relative depth would change its acceleration. **

This is an entertaining thought experiment, but it’s not easy to think this way.

GA

GingerAle Jul 13, 2020