If block A is moving horizontally with velocity `v_(A)`, then find the velocity of block B at the instant as shown in fig.

Options

A. `(hv_(A))/(2sqrt(x^(2)+h^(2)))`
B. `(xv_(A))/(sqrt(x^(2)+h^(2)))`
C. `(xv_(A))/(2sqrt(x^(2)+h^(2)))`
D. `(hv_(A))/(sqrt(x^(2)+h^(2)))`

Answer

C

Explanation

From fig., `L=2h-2y+sqrt(x^(2)+h^(2))`
Differentiating the equation, we get
`(dy)/(dt)=(x)/(2sqrt(h^(2)+x^(2))) (d)/(dt)`
`implies V_(B)=(xV_(A))/(2sqrt(h^(2)+x^(2)))`