Aball is thrown from a height of 157 feet with an initial downward velocity of 9/fts . the ball's height h (in feet) after t seconds is given by the following. =h−157−9t-16tsquared how long after the ball is thrown does it hit the ground? round your answer(s) to the nearest hundredth. (if there is more than one answer, use the "or" button.)

  2.86 seconds

Step-by-step explanation:

We presume the height is measured above the ground, so the height when the ball hits the ground is 0 feet. Putting that into the equation and solving for t, we get ...

  0 = 157 -9t -16t^2

We can use the quadratic formula with a=-16, b=-9, c=157 to find the solution.

  t = (-b -√(b^2 -4ac))/(2a) . . . . . we are not interested in the negative solution

  t = (-(-9) -√((-9)^2 -4(-16)(157)))/(2(-16))

  t = (9 -√10129)/-32 = (9 -100.6429)/-32 ≈ 2.86384

The ball will hit the ground about 2.86 seconds after it is thrown.

for the first question it would be the second answer for the second question it would be the third answer and for the third question it would be the first answer

step-by-step explanation:

step-by-step explanation:

because 1/3 is a fraction. we can tell because there is a divisor symbol between the 1 and the 3. if we were to add 2 fractions together, we would add the numerators, which is the number on top, but not the denominators, which is the number on the bottom

1/3+1/3=2/3

hope this !