A ball is dropped from rest 3.0 m directly above the vertex of a concave mirror that has a radius 1.0 m and lies in a horizontal plane.(a) Describe the motion of ball's image in the mirror (i.e,for what positions image is real and when it is virtual..)(b) At what time do the ball and its image coincide?

1. As the ball approach the mirror, just before reaching the centre of curvature, C of the mirror, a smaller real image of the ball will be seen moving away from the mirror, formed between principal focus, F and C.

When the ball is just 1 m above the mirror, its image is also formed at the same location, that is 1.0m above the mirror, as the ball further moves towards the mirror, anywhere between F and C, a smaller real image can be seen rising vertically upwards, at a position greater than 1.0m above the mirror.

Further more, as the ball further descends down towards the mirror, when it is between the pole, P and F, a large virtual image of the ball will be observe to be formed at the back of the mirror.

(2) The ball will coincide with its image when it has fallen 2.0m. Using the equation s=(1/2)gt^2, where s=2.0m, g=9.8m/s^2, then, get

                t=0.6388 s = 0.64 s Ans

Initially, 

        u = -3m

        v = ?

        f = -0.5m

Now, using mirror formula 

        1/f = 1/v+1/u

        v = -0.6m

 magnification(m) = -v/u

        m = -1/5 

   i.e. image is real, 1/5 times smaller than object and inverted

The image moves away from mirror as object comes near to it and the image becomes virtual when object comes between focus and pole i.e. less than 0.5 from mirror

(b) The object and image coincides at center of curvature i.e. 1m from mirror

     by the equation of motion

     s = 1/2gt²

    s = 2m & g = 9.8m/s²

   so, t = 0.639sec