tag:blogger.com,1999:blog-5997147895880585179.post5002372735735568194..comments2024-03-17T18:54:13.159+05:30Comments on Solutions to I E Irodov - Physical Fundamentals of Mechanics: Irodov Problem 1.27Unknownnoreply@blogger.comBlogger2125tag:blogger.com,1999:blog-5997147895880585179.post-48032526385537945782011-06-12T10:23:24.452+05:302011-06-12T10:23:24.452+05:30This is also a valid way of doing it :-)This is also a valid way of doing it :-)Krishna Kant Chintalapudihttps://www.blogger.com/profile/07024636005277277405noreply@blogger.comtag:blogger.com,1999:blog-5997147895880585179.post-76040409502672534852010-12-12T19:59:42.219+05:302010-12-12T19:59:42.219+05:30i have got another solution without using calculus...i have got another solution without using calculus<br />please tell me whether my aproach is correct or not<br />y=ax-bx^2<br />let the horizontal component of its velocity at the origin be Vx and vertical component of velocity at the origin be Vy<br />as the acceleration is along the negetive y direction<br />motion along the x direction is uniform. acceleration=-ω<br />eqn on motion along x direction <br />x=Vxt<br />along y direction<br />y=Vyt-1/2ωt^2<br />we know that<br />y=ax-bx^2<br />substituting for x we get y=a(Vxt)-b(Vxt)^2<br />equating we get<br />Vx=aVy<br />bVx^2=1/2ω<br />using these eqns to fing root of(Vx^2 + Vy^2)<br />which is the velocity at origin<br />= root of [(1+a^2)ω/2b ]Anonymousnoreply@blogger.com