The National Football League is holding Superbowl 45 in the Dallas Cowboy stadiu

Question

The National Football League is holding Superbowl 45 in the Dallas Cowboy stadium. One concern is the possibility of kickoffs or punts striking the video center that hangs from the roof. A quote from an architectural press release on the new stadium is as follows.

“Hanging approximately 90 feet above the field from the roof structure, the innovative video center spans between the 20-yard lines and features four individual boards – two facing the sidelines and two facing the end zones.”

The maximum distance a projectile travels over horizontal ground, neglecting air resistance, is obtained if the initial projectile velocity is at 45° to the horizontal. Assuming punt is made at a 45° angle and just misses the video board, what is the maximum horizontal distance the punt will travel.

Explanation / Answer

We can separate the side-to-side motion from the up-and-down motion. The sideways progress of the ball keeps going at constant rate down the field and we can use D = RT. The up-and-down motion has gravity, so the ball slows to zero speed at its maximum height and then speeds up on its way back to the field. To go up 90' requires a certain initial velocity component in the vertical direction. It looks like you are using English system units. Since we can neglect wind resistance we use one of the kinematic equations (Eqn 1) Vf^2 = Vo^2 +2(xf - xo)a where Vf is the final velocity, Vo is the starting speed, xf is the final height, xo is the starting height. Now, what comes up must come down, so upward component of the starting velocity is going to equal the downward component of the final velocity (the field is flat, so the ball goes up ninety feet and down ninety feet). If we substitute these numbers into (Eqn 1) we get 0 = 0. This isn't useful. A nice trick is to break the flight in half. At 90', the upward velocity is zero if the ball is to just skim the video center. Now we can find the starting velocity. Vo^2 = -2(xf - xo) a Vo^2 = -2(90 ft)(-32 ft/s^2) = 5760 ft^2/s^2 Vo = 75.9 ft/s Since the sides of a 45-45-90 right triangle are equal, the horizontal component of the velocity is also 75.9 ft/s. Now, let us find out how long it takes the ball to go up 90' (or fall from 90' up down to the ground) (EQN 2) Vf = Vo + at, where t is time. We could cut the vertical flight in half and then double to get the time the ball was moving down the field, or we could solve for the entire vertical flight. Lets solve for the entire vertical flight. Let's call up + and down -. t = (Vf-Vo)/a = (-75.9ft/s - +75.9 ft/s)/(-32 ft/s^2) = 4.74 s (we neglect the height of the ball above the ground when it was punted) The ball is in the air for 4.74s. This is also the time that the ball travels down the field at constant horizontal speed. D = RT = 75.9 ft/s(4.74s) = 360 ft = 120 yards. If someone could punt from within his endzone and have the other team get a touchback, the video center could be a problem with a 45 degree kick and no wind resistance, but no one punts that far. With wind resistance, a lower angle gives more range, so the real life situation is even less of a risk.

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