smokinHawk1647545499 Posted September 19, 2005 Report Share Posted September 19, 2005 ok its an almost standard projectile motion problem. trying to figure it out, hoping that its not some simulanious equation problem or something, the formula to solve alludes me must be some simple way to do it. projectile is launched at an unknown velocity at an 45* angle. it reaches its target at Horizonal distance it makes is 70ft. Verticle distance it makes is 6ft above launch point. I need to find the intial velocity and time of flight. thanks Quote Link to comment Share on other sites More sharing options...
smokinHawk1647545499 Posted September 19, 2005 Author Report Share Posted September 19, 2005 the formula i came up with to find time is (2*(verticle distance - horizontal distance)/gravity)^(1/2) since its at a 45* angle: velocity horizontal = velocity verticle anyone want to confirm if those are right? Quote Link to comment Share on other sites More sharing options...
Berto Posted September 19, 2005 Report Share Posted September 19, 2005 is the target at the same level as the lauch point? your assumption of vertical speed equal to horizontal speed is correct because sin(45) and cos(45) are the same value. the relationship between time of flight and Vo (initial velocity) is easy since you have the constant acceleration of gravity. You can assume that the horizontal velocity remains unchanged because drag effects are negligible (for your class). so then if you know how far the horizontal target is you can get the initial horizontal velocity. and by rule the vertical velocity is the same. i might have biffed the explanation, i really like to see drawings of problems :-/ Quote Link to comment Share on other sites More sharing options...
smokinHawk1647545499 Posted September 19, 2005 Author Report Share Posted September 19, 2005 No the target is 6 feet higher then the launch point something seems to be wrong with my above equation as the numbers arent jiving when i put them back into different equations. Quote Link to comment Share on other sites More sharing options...
Trouble Maker Posted September 20, 2005 Report Share Posted September 20, 2005 I'll do it tomorrow, sleep now (physics minor). Quote Link to comment Share on other sites More sharing options...
Lustalbert Posted September 20, 2005 Report Share Posted September 20, 2005 ADD keeps me from fully enjoying math like you guys. Quote Link to comment Share on other sites More sharing options...
smokinHawk1647545499 Posted September 21, 2005 Author Report Share Posted September 21, 2005 anyone figure this out, i have gotten no where. it cant be that hard, its not like im asking you to define what causes gravity. Quote Link to comment Share on other sites More sharing options...
black00ws6 Posted September 21, 2005 Report Share Posted September 21, 2005 v sin 45 = vo, substitue v sin 45 for vo in the equation of 2ad = vf^2 - vo^2 2 * -9.8 m/s^2 * 1.83 m = (0 m/s)^2 - (v sin 45)^2 -35.87 m^2/s^2 = -(v sin 45)^2 35.87 m^2/s^2 = (v sin 45)^2 sqrt (35.87 m^2/s^2) = v sin 45 5.99 m/s = v sin 45 v = (5.99/sin 45) m/s v = 8.47 m/s To find the time.... 8.47 m/s sin 45 = 5.99 m/s , y component t * 5.99 m/s = 21.34 m t = 3.56 s I'm sure I f'ed up something.. been about 8 years since doing this crap, jesus, i'm old Quote Link to comment Share on other sites More sharing options...
smokin5s Posted September 21, 2005 Report Share Posted September 21, 2005 Jesus John... Quote Link to comment Share on other sites More sharing options...
black00ws6 Posted September 21, 2005 Report Share Posted September 21, 2005 It's not right, I think i have a conversion off somewhere.. The time and x component should equal the distance traveled. should be right now.... SHOULD be Quote Link to comment Share on other sites More sharing options...
smokinHawk1647545499 Posted September 22, 2005 Author Report Share Posted September 22, 2005 thanks still something looks off, ill go over your figures and check to see if i see something, might be right though i dont know. Quote Link to comment Share on other sites More sharing options...
black00ws6 Posted September 23, 2005 Report Share Posted September 23, 2005 so was it right? Quote Link to comment Share on other sites More sharing options...
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