Aball is kicked upward with an initial velocity of 56 feet per second. the ball's height h (in feet) can be expressed as a function of time t (in seconds) by the equation h = -16t2 + 56t. how much time does the ball take to return to the ground?
a) 1.75 seconds
b) 2.50 seconds
c) 2.75 seconds
d) 3.50 seconds

  D)  3.50 seconds

Step-by-step explanation:

You are interested in the positive value of t such that h=0. Substituting 0 for h, we get ...

  0 = -16t^2 +56t

  0 = -16t(t -3.5) . . . . . factored form

The product is zero when the factors are zero. Here, the factor of interest is ...

  0 = t-3.5

  3.5 = t . . . . . add 3.5 to both sides

The ball takes 3.5 seconds to return to the ground.

let's first notice, the center of the hyperbola is at the origin, since the focus points are equidistantly 3 units away from the origin, that also means that the distance "c" from the center to either focus is 3.

it should also be notable that, the traverse axis for this hyperbola is vertical, or namely over the y-axis and therefore, the positive fraction will be the one with y-variable.

the traverse "a" component or distance from the center to either vertex is not 2, is more than 2, so we can safely put it at √5 ≈ 2.24.