yes it is v/sin theta
A block is dragged on a smooth plane with the help of a rope which moves with velocity v shown in Fig. The horizontal velocity of the block is
a)v
b) vsinÎ¸
c) v/sinÎ¸
d) v/cosÎ¸

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well from triangle we can say x=√y^{2}h^{2}
so, dxdt=2y2√y^{2}h^{2}.dydt
also dydt=v
so v_{x}=v.(y√y^{2}h^{2})=v.yx=vsin Î¸
sin Î¸= yx
can anyone tell me that how can we approach that it is a type of constraint relation.
i want to know the basic approach
the basic approach is v=ds/dt wer s= displacement
and a=dv/dt
thus if we can some how relate the displacements of the particles we can relate their velocities and accelerations!! :)
i easily explained yesterday to my friend yesterday that answer is v/sintheta but today after 22 hrs i forgot myself that how my answer came yesterday
The component of block's velocity along the rope is equal to v.
v'sin(theta)=v (v' is velocity of block)
this is due to the fact that the string is unstretchable this can be proven by constaint relations or power method.