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# help asap

 P: n/a HELP ME PLEASE!! my email is th**************@yahoo.com I can't get the ball to go up right side and then I need it to turn around and keep turning until velocity=0 I have been at it for the past 2 weeks now i give up and call for help. Please if anyone can gide me through i will be so grateful!! I have pasted my code below from cmath import * from visual import * floor1 = box(length=10, height=0.5, width=4, color=color.blue) floor1.pos = (-6,4,0) floor1.axis= (5,-5,0) floor2 = box(length=10, height=0.5, width=4, color=color.blue) floor2.pos = (6,4,0) floor2.axis= (-5,-5,0) floor3 = box(length=7, height=0.5, width=4, color=color.blue) floor3.pos = (0,1.25,0) ball= sphere(radius=0.5, color=color.red) ball.pos=(-8.6,7.5,0) m=3. #kg angle=asin(3.6/5.)#radians g=-9.8 mu=.2 N=mgcos(angle) F=mgsin(angle) f=mu*N lax=(-Nsin(angle)+fcos(angle))/m lay=(-Ncos(angle)-fsin(angle)+m*g)/m rax=(+Nsin(angle)+fcos(angle))/m ray=(-Ncos(angle)-fsin(angle)+m*g)/m ds=0.01 dt=0.01 vx=lax*dt vy=lay*dt ball.velocity=vector(vx,vy,0) #print a while 1: rate(100) ball.velocity.x=ball.velocity.x+lax*dt ball.velocity.y=ball.velocity.y+lay*dt ball.pos=ball.pos+ball.velocity*dt if ball.x>-3.5 and ball.x<=3.5: vx=sqrt(2(-gball.y+0.5(vx2+vy2)-fds)) ball.velocity.x=ball.velocity.x+mugdt ball.velocity.y=0 ball.pos=ball.pos+ball.velocity*dt if ball.x>3.5 and ball.x<8.6: vx=vx*cos(angle) vy=sqrt(2/m(m-gball.y-fds)) #print vy vy=vy*sin(angle) #print vy ball.velocity.x=ball.velocity.x+rax*dt ball.velocity.y=ball.velocity.y+ray*dt ball.pos=ball.pos+ball.velocity*dt #print ball.pos Jul 18 '05 #1 