1. The problem statement, all variables and given/known data A rod AB is shown in figure. End A of the rod is fixed on the ground. Block is moving with velocity √3 m/s towards right. The velocity of end B when rod makes an angle of 60^{o} with the ground. 2. Relevant equations 3. The attempt at a solution I assumed that at any instant the distance of A from block is x and distance of B from ground is y. The length of rod is l. l^2=x^2+y^2 Differentiating with respect to time. 0=2x(dx/dt)+2y(dy/dt) y=xtan(60^{o}) dx/dt=√3 m/s Solving, i get (dy/dt)=-3 m/s But that's not correct, the answer says its 2 m/s. I don't seem to get where i am wrong. Any help is appreciated. Thanks!
Hi Pranav! Are you sure about that step? Recheck your solving, [tex]0 = 2x\sqrt{3} + 2\sqrt{3} x \frac{dy}{dt}[/tex]