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MX, SX & Off-Road Discussions
General Moto | Off-Topic Posts
how big of jump
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[QUOTE="Yz426King, post: 164344, member: 24563"] Here is a distance formula for you: d= distance traveled v^2=velocity in feet per second squared a=angle of projectile d = (v^2/32)(sin(2a)) so if the angle is 45 deg. and the speed is 35Mph: d = (51.3^2/32(1) d = 82.347 The Maximum distance that is physicaly possible at the stable velocity of 35 Mph is 82 ft. (That is if you are on earth [gravity] , the wind is not blowing, and you are not accelerating or decellerating.) Here is another for height: (ill let you work it yourself! :eek: :eek: ) y=height x=horisontal distance a=angle h=height from ground level from which you are shooting (ramping) the object. v=velocity in ft/sec y = -(16/(v^2*(cos*a)^2)*(x^2 + (tan*a)x + h) FYI this materiel was taken directly form an algebra2 book. [/QUOTE]
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MX, SX & Off-Road Discussions
General Moto | Off-Topic Posts
how big of jump
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