What  is the mass of the soccer (football)ball??
409505
A soccer ball of mass 0.43 kg leaves the foot of the kicker with an initial speed of 25 m/s.The weight of the ball at the start of the game shall not be more than 450 g nor less than 410 g.\nSource-Sam
239851
According to your question statement, the pressure inside the soccer ball is less than that of the surrounding atmosphere. This seems suspect...
875895
Find someone with a soccer or rugby ball (local football club?) - it will more than likely have the same fitting. \n\nEven if they haven't got one they will probably know of somewhere you can get it inflated.\n\nFailing that - call your local gym and see if they can do it for you.
566517
soccer ball: \nhttp://images.google.com/images?hl=en&q=soccer%20ball&sa=N&tab=wi\n\nsoccer (general): \nhttp://images.google.com/images?svnum=10&hl=en&lr=&q=soccer\n\nstatue of liberty:\nhttp://images.google.com/images?svnum=10&hl=en&lr=&q=statue+of+liberty
893929
Fg = mg = (0.20 kg)(9.80662 m/s^2) = 1.961324 N\nFc = (mv^2)/r = mrω^2\nFc = (0.20 kg)(0.60 m)((2π rad/s)^2\nFc = 4.73741 N\nFt = √( 4.73741^2 + 1.961324) @ arctan(1.961324/4.73741)\nFt ≈ 5.127 N @ 22.5°\n\n[The vertical component of Ft must offset the weight of the ball.]
393333
there are no spots on any soccer ball, they are pentagons and hexagons.  Spots are round.
291257
Well silver density is 10.5 Kg/dm³. but rugby ball volume is quite difficult to calculate... I made an aproximation calculated as two opposite cones taking the max. size and it gave me a little more than 3 dm³, so the wheight would be about 31.5 Kg. Actually even more as the ends of the cones in the ball are rounder so they have a bigger volume...
987577
FIFA World Cup Trophy\nThe replacement trophy, officially known as the FIFA World Cup Trophy, was first presented at the 1974 World Cup. Designed by Silvio Gazzaniga and produced by Bertoni, Milano, it stands 36.5 cm (14.4 in) tall and is made of 5 kg (11 lbs.) of 18 carat (75%) solid gold with a base (13 cm [5.1 in] in diameter) containing two layers of malachite. The trophy, which weighs 6.175 kg total, depicts two human figures holding up the Earth.
737095
soccer because I never heard of speed ball
308702
A soccer ball(played in a FIFA match) measures between 27 and 28 inches in circumference and weighs between 14 and 16 ounces.\nSource-SAMM
1010100
ok, so if we're assuming that it was a head on elastic collision, and all the energy was transferred then there are two possible situations: if the 25lb ball was sliding then the total energy transferred will be equal to the amount it had originally (since we assumed elastic collision) that is equal to 1/2mv^2...where mass would usually be measured using the metric system. \nif the ball was rolling then it has not only translational energy (1/2mv^2) but also rotational energy which is equal to 1/2Iw^2 where I = inertia of a ball...if it is a solid sphere the moment of inertia is equal to 2/5mr^2 (m = mass, r = radius) and w (omega) = rotational speed in radians per second...which is equal to velocity/radius. so the amount of energy transferred would be equal to 1/2Iw^2 + 1/2mv^2.
833406
The world cup ball.Trust me!!!!!!!
1033468
2.36 m s^-1 to 2 d p\n\nWe split the launch velocity of the ball into a vertical and a horizontal component.\nHorizontal component = 19.8 * cos(49.1) = 12.9639 m/s\nVertical component = 19.8 * sin(49.1) = 14.9659 m/s\n\nGravity accelerates the ball vertically with the acceleration -9.8 m s^-2 (negative because the pull of gravity is downwards while the ball is moving upwards to start with)\n\nThe vertical distance from the ball to the ground after time t is given by the formula s = u * t + (g * t^2) / 2 where: u = initial velocity, g = acceleration of gravity, t = time elapsed\n\nWhen the ball returns to the ground, s = 0\nTherefore the time taken for the ball to return to the ground is given by\nu * t + (g * t^2)/2 = 0 Substitute values\n14.9659 * t - (9.8 / 2) * t^2 = 0 Remove brackets, divide by t, rearrange\n4.9 * t = 14.9659\nt = 3.05427 s\n\nDuring that time the ball will travel horizontally \ns = u * t where s is distance in m, u is velocity in m s^-1, t is time in sec\nDistance travelled horizontally is 12.9639 * 3.05427 = 39.5953 m\n\nThe player has to run 46.8 - 39.5953 m = 7.20475 m in 3.05427 s\nHis speed must therefore be 2.35891 m s^-1
1039271
Why would any sane person shoot a soccerball? When you shoot them they pop.
433746
An appendix to the "Rules of Golf" defines that a golf ball must not weigh more than 45.93 grams, that its diameter must not be less than 42.67 mm, and that its shape may not differ significantly from a symmetric sphere. Like golf clubs, golf balls are subject to testing and approval by the Royal and Ancient Golf Club of St Andrews and the United States Golf Association, and those that do not conform with the regulations may not be used in competitions (Rule 5-1).
1157509
The ball is: \nspherical; \nmade of leather or other suitable material; \nof a circumference of not more than 70 cm (28 ins) and not less\nthan 68 cm (27 ins); \nnot more than 450 g (16 oz) in weight and not less than 410 g\n(14 oz) at the start of the match; \nof a pressure equal to 0.6 – 1.1 atmosphere (600 – 1100 g/cm2) at\nsea level (8.5 lbs/sq in – 15.6 lbs/sq in).
901812
For a little joke: Find a truncated icosahedron.  The scavengers can fulfill this requirement by bringing back a soccer ball.