Dr. Dazzle Posted October 4, 2005 Share Posted October 4, 2005 Alright, so I'm taking highschool physics since I didn't take it in highschool, and there's a couple questions on this assignment that I can't seem to solve. Now these are pretty simple questions but since I'm the most retarded person when it comes to math/science/physics, I turn to you, oh wise 12oz nerds, to lend a helping hand. This will probably turn in to a regular thing, since I have a whole semester of this shit. 1. Ideal projectile problem where the initial and final heights are the same. Consider a projectile launcher which fires an object at an angle of 41.0 degrees at a velocity of 10.2 m/s. What is the horizontal distance travelled and velocity just before it lands? I've solved for the V initial in the Y and X directions. I think the whole intial and final heights being the same thing is throwing me off. Does that mean that the displacement in the Y direction is cancelled out?What formula would I use? This is the information I've got for the Y direction: a = -9/81 m/s^2 d = ?? Vo = 10.2 m/s Vf = ?? t = ?? I think I know how to solve for Vf and time, so do I do that and then solve for d? 2. A cannon at an angle of 35 degrees fires a cannon ball through a hoop that is 19.6 m away at a height of 2.3 m. What is the initial velocity and the maximum height the ball will reach? This one I'm just plain fucked with. How can you find the X and Y components without the Vo? Yeah any help would be greatly appreciated. Sorry if these are easy questions, I told you, I'm an idiot. Thanks. I like how the one time I actually need 12oz, it's down. Awesome. Quote Link to comment Share on other sites More sharing options...
Xeroshoes Posted October 4, 2005 Share Posted October 4, 2005 1. Solve for the vertical and horizontal components of initial velocity, using the angle and your knowledge of trigonometry. The acceleration due to gravity acts only on the vertical component, so the horizontal component remains constant. Since the final height is the same, the final velocity will also be the same, but the vertical component will be directed downward rather than upward. Now you solve for the time. The total time is simply the time it takes for the acceleration due to gravity to reduce the vertical component to zero, then back to the same speed but in the opposite direction. That is, at = 2*Vo(Vertical Component.) Solve this for t. Once you have t, simply multiply it by the horizontal component of the initial velocity, and you will have the horizontal distance traveled. I think that's all correct. Quote Link to comment Share on other sites More sharing options...
!@#$% Posted October 4, 2005 Share Posted October 4, 2005 effing crashes that's good advice physics is all about problem solving and being able to make educated guesses and assumptions get ready to figure shit out, that a formula will not tell you and the problem doesn't make obvious (such as the initial/final heights and velocities being the same, factoring in for the gravity in some problems as acceleration, and angles of projectiles) Quote Link to comment Share on other sites More sharing options...
Kr430n5_666 Posted October 4, 2005 Share Posted October 4, 2005 call me or not not me or call not call or me me not or call me call or not Quote Link to comment Share on other sites More sharing options...
Fondles Posted October 4, 2005 Share Posted October 4, 2005 mmmmm....call u. Quote Link to comment Share on other sites More sharing options...
Future Droid Posted October 5, 2005 Share Posted October 5, 2005 using v(initial), dont forget to account for gravity, break it down into x and y components. figure out where the top of the arch is (x and y). your total distance will be 2y the final velocity should be found using x(top of arch), forget about y, as there is no gravity in the y direction, and solve for v(final). i think thats right..... Quote Link to comment Share on other sites More sharing options...
Dr. Dazzle Posted October 5, 2005 Author Share Posted October 5, 2005 Originally posted by Future Droid@Oct 4 2005, 05:50 PM using v(initial), dont forget to account for gravity, break it down into x and y components. figure out where the top of the arch is (x and y). your total distance will be 2y the final velocity should be found using x(top of arch), forget about y, as there is no gravity in the y direction, and solve for v(final). i think thats right..... Quoted post It's the other way round. Gravity goes in the Y direction. I'm going to get up early tomorrow to do this assignment. Thanks for the help so far guys, if anyone else wants to add by all means do. Quote Link to comment Share on other sites More sharing options...
Future Droid Posted October 5, 2005 Share Posted October 5, 2005 yeah my bad, too many drugs last night...duhhh and distance 2x. Quote Link to comment Share on other sites More sharing options...
Abracadabra Posted October 5, 2005 Share Posted October 5, 2005 why on earth did i come into this thread? Quote Link to comment Share on other sites More sharing options...
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