263a=(mg-kx)/m............(1)[where x is the extension of the string after reaching it's natural length]
Now at the point where the string regains it's natural length velocity is √2gl.See from the equation (1) that the acceleration is positive if x<=mg/k.
now in equation (1) writing a as v(dv/dx),we get
v(dv)=(g-kx/m)(dx).................(2)
Integrating this equation with the boundary condition on v from V to √2gl and on x from mg/k to 0 we get V2=2GL + mg2/k.This is the maximum velocity of the block because till x=mg/k acceleration is positive.
Now after x=mg/k acceleration is negative.The block's velocity reduces from V to 0.Now it is already given that the block touches the ground when it's velocity is 0.So integrate again equation (2) with the conditions that :
Lower limit of v=V
Upper limit of v=0
Lower limit of x=mg/k
Upper limit of x=h,where h is measured from the point where the string attains it's natural length.
A quadratic equation comes in term of h.Solve it to reach the answer.
Also plz tell the answer to question (3).
