181k views
0 votes
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.

1 Answer

5 votes
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
User Yart
by
8.8k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories