Question

A particle is dropped along the axis from a height f/2 on a concave
mirror of focal length f as shown in the figure. The acceleration due
to gravity is g. Then the maximum speed of the image is given by
hxfg where x =
pole P​

Answer 1

Hence the maximum speed of the image is V I (max) = 3 / 4 √ 3 f g

Explanation:

v(OM) = − m^2

m = f / f - u

m = - f / - f + ( f / 2 - ft^2 / 2 )

m = 2 f / f + greater than^2

v1 =  - ( 2 f / f + greater than )

v1 = - 4 f^2 / ( f + greater than^2 )^2

Maximum speed = dv / dt = 0

t = √ f / 3 g

V I (max) = 3 / 4 √ 3 f g

Hence the maximum speed of the image is V I (max) = 3 / 4 √ 3 f g