The equation y = -16t^2 + 28t + 180 describes the height (in feet) of a ball tossed up in the air at 28 feet per second from a height of 180 feet from the ground. In how many seconds will the ball hit the ground?
+2666 We can set y to 0, and solve for this quadratic.
This will yield: \(0=-16t^2+28t+180\)
We now have a quadratic, and can use this formula: \(t = {-b \pm \sqrt{b^2-4ac} \over 2a}\) to solve for t, because the equation is in the form \(ax^2+bx+c=0\).
Can you take it from here?
