A lens with f= 20.0 cm creates a virtual image at -37.5 cm (in front of
the lens). The object is 4.44 cm
tall. How tall is the image?
(Mind your minus signs.)
(Unit = cm)

h ’= 12,768 cm

Explanation:

For this exercise let's use the constructor equation

         1 / f = 1 / p + 1 / q

where f is the focal length, p the distance to the object and q the distance to the image

the magnification equation is

          m = h '/ h = -q / p

let's find the distance to the object

         1 / p = 1 / f- 1 / q

         1 / p = 1/20 - 1 / (- 37.5)

         1 / p = 0.076666

          p = 13.04 cm

now let's use the magnification equation

         h ’= - q / p h

let's calculate

         h ’= - (-37.5) / 13.04 4.44

         h ’= 12,768 cm

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer