first name one pulley as A and other as B

and let midpoint is C

now in triangle OAC....where O is a point at m.......

l^2=y^2+a^2

where l is length of string from A to O

and Y is oc

A is AC

now differentiate w.r.t. t

u get 2ldl/dt =2ydy/dt+0

dy/dt = dl/dt *l/y =(dl/dt)/(y/l)

now suppose dl/dt = u

and y/l = cos(theta)....from fig....

now dy/dt=u/cos(theta)

i.e. upward speed of mass m is u/cos(theta)

.....

i want to know as mass is pulling upwards..from both sides so the vel. should be...2u/cos(theta) ,,,but i got u/cos(theta)....from constraint law....

plzzzzz help;;;;;

 

velocity. 

Now since the angle between the string and the block is 

Angle @ 

Now see the length l of string decreases at the rate of 

U m/s. 

2L dL/dT = 0 + 2y dy/dt 

differentiation of b² = 0 as it remains constant. 

now we get  

dy/dt = l(dL)/ y(dt)