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Calling all math brains: I need an equation to account for somebody jumping off a roof to land on a moving car
I'm trying to finish the last details on my book, and I need a quasi-complicated equation to account for a slim, but very densely constructed female (about 280 pounds), leaping off a roof (between six and twelve stories), aiming for a small moving car (travelling at about 65 kph).
All details are negotiable, and better suggestions are welcome.
I need the speed and rate she would fall at, and how far away the car would have to be when she jumped, accounting for jump time and windshear, etc.
I have no mind for math, so any help would be appreciated. Any insomniac physics wizards out there tonight...?
I suppose I should have added the stipulation, 'Only real answers, please.'
The weight of the woman is irrelevant--gravity accelerates all mass at 9.8 m/sec2. The time the object (person) takes to fall, and the time of the moving car must be equal. So, for the person, t=(2s/g)^1/2 , where s is the height of the leap, and g is gravity (9.8 m/sec2) For the car, t=d/r, where d is how far away the car is, and r is its speed (65 kph, which needs to be changed to m/sec by multiplying 1000m/k and dividing 3600 secs in an hour). so the equation is formed by setting the two times equal and substituting. you arrive at (2s/g)^1/2=d/r. Transpose this equation to get the distance the car must must travel--d=r(2s/g)^1/2. s=height of leap in meters (lets say 9 stories or 35 meters). r=speed of car which is 18 m/sec. g is 9.8 m/sec2. So d=(18)[(2)(35)/9.8]^1/2. The car should be 48 meters from the female's landing point under these conditions.