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.
When the ball hits the ground, y will be = zero
zero 0 = x^2-8x-33 solve for x .... throw out any negative values
just to double check is x=11, -3
if it is do i add them?
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.