SOLUTION: The force exerted on the soles of the feet for a typical basketball jump shot is directly proportional to the shooter's weight. If a 150 lb player exerts a force of 2000 lbs, what

Question 296682: The force exerted on the soles of the feet for a typical basketball jump shot is directly proportional to the shooter's weight. If a 150 lb player exerts a force of 2000 lbs, what force is exerted by a 6ft 8in. 280 lb player?
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Let F = force exerted and w = weight

So "The force exerted on the soles of the feet for a typical basketball jump shot is directly proportional to the shooter's weight" means that for some constant 'k'

Because we're given that "a 150 lb player exerts a force of 2000 lbs", we know that and . Plug these into the equation above to get . From here, divide both sides by 150 to get

So the equation is . Now just plug in to find 'F' for a 280 lb player.

Note: the given info on the height of the player has nothing to do with the problem. That's just there to throw you off.
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