An inextensible string AB is tied to a block B of negligible dimensions and passes over a small pulley C so that free end A hangs h unit above the ground on which the block B rests. In this initial position shown in fig. The free end A is h unit below C. if now end A moves horizontally with a velocity `v_(0)`, obtain an expression for the velocity of the block at any time t.

Explanation

In time t, the end A moves to the position `A_(1)` so that distance `A A_(1)=v_(0)t`. The block B moves upwards to the position `B_(1)`. Let `BB_(1)=y`.
Then length `A_(1)C=h+y`
In the `Delta ACA_(1), A_(1)C^(2)=AC^(2)+A A^(2)`
`(h+y)^(2)=h^(2)+(v_(0)t)^(2)`
or `h^(2)+y^(2)+2hy=h^(2)+v_(0)^(2)t^(2)`
or `y^(2)+2hy-v_(0)^(2)t^(2)=0`
After solving eq. (i) we get
`y=-h+sqrt(h^(2)+v_(0)^(2)t^(2))`
This is the equation for the displacement y of the block.
For velocity of the block differentiating equation (i) w.r.t. time
`v=(dy)/(dt)=(d)/(dt)[-h+(h^(2)+v_(0)^(2)t^(2))^(1//2)]`
`=1/2 (h^(2)+v_(0)^(2)t^(2))^(-1//2) xx 2 v_(0)^(2)t`
`=(v_(0)t)/((h^(2)+v_(0)^(2)t^(2))^(1//2))`.