Question

A rubber ball is held 4.0 m above a concave spherical mirror with a radius of curvature of 1.5 m. At time equals zero, the ball is dropped from rest and falls along the principal axis of the mirror. How much time elapses before the ball and its image are at the same location?

Answer 1

The time elapsed when the ball placed above the concave mirror and the image formed would be at the same location is 0.55 s.

Image distance

The position of the image formed is determined using the followimg mirror formula;

$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$

where;

• f is the focal length of the mirror
• v is the image distance
• u is the object distance

f = R/2

f = 1.5/2

f = 0.75 m

When the ball and its image is in the same position, u = v

The position of the ball is calculated as;

$\frac{1}{f} = \frac{1}{u} + \frac{1}{u} \\\\\frac{1}{f} = \frac{2}{u} \\\\u = 2f\\\\u = 2(0.75)\\\\u = 1.5 \ m$

Time of motion of the ball

The time taken for the ball to travel the caluclated distance is determined as;

h = ut + ¹/₂gt²

1.5 = 0  + ¹/₂(9.8)t²

1.5 = 4.9t²

t² = 1.5/4.9

t² = 0.306

t = 0.55 s

Thus, the time elapsed when the ball and its image are at the same location is 0.55 s.

Learn more about concave mirror here: brainly.com/question/7512320