Former nfl punter ray guy holds the record for the longest hangtime on a punt. if the ball leaves with an upward velocity of 128 ft/s from an initial height of 4 feet, how long will the ball be in the air? use the formula h= -16t(squared) +128t +4, where h is the height of the ball in feet and t is the time in seconds since it is kicked. round your answer to the nearest tenth.

Since we have the formula for height, all we need to do is solve for t when the height of the ball is 0, meaning it is on the ground. The equation is H = -16t^2 + 128t + 4 Substituting H for 0, we get: 0 = -16t^2 + 128t + 4 Now the problem becomes a simple quadratic equation that we can solve using the quadratic formula. The quadratic formula for at^2 + bt + c = 0 is: t = [-b +/- sqrt(b^2 - 4ac)] / 2a Plugging in a, b, and c, we get t = [-128 +/- sqrt(128^2 - 4*-16*4)] / -32 Solving for t, we get t = 8.03, -0.03. Since the time must be positive, the answer is 8.0. The ball is in the air for 8.0 seconds