A particle is dropped along the axis from a height `(f)/(2)` on a concave mirror of focal length f as shown in figure. Find the maximum speed of image .

Explanation

`v_(M1) = -m^(2)v_(OM) = -m^(2)(gt)` where.
`m = (f)/(f-u) = (-f)/(-f + ((f)/(2) - (ft^(2))/(2))) = (2f)/(f +gt^(2)) implies v_(1) = -((2f)/(f+ gt^(2)))^(2)(gt) = -(4f^(2)gt)/((f + gt^(2))^(2))`
For maximum speed `(dv_(I))/(dt) = 0 implies t= sqrt((f)/(3g)) implies v_(Imax) = (3)/(4)sqrt(3fg)`