According to the question;

Object distance (u) = -60cm;

Image distance (v) = 120cm;

Focal length = f

By lens formula;

\(\frac1v\,-\,\frac1u\,=\frac1f\)

⇒ \(\frac1{120}\,-\,\frac1{-60}\,=\frac1f\)

⇒ \(\frac1{f}\,=\,\frac1{120}\,+\frac1{60}\)

⇒ \(\frac1{f}\,=\,\frac{1+2}{120}\,+\frac3{120}\)

⇒ f = \(\frac{120}3\)

⇒ f = 40cm.

Therefore Focal length of lens is 40 cm.

Since the focal length is positive. Therefore the lens is convex lens of focal length 40cm.

Now;

Height of object h_{1}= 5cm;

Magnification =\(\frac{h_2}{h_1}\) =\(\frac{v}u\)

Putting values of v and u

Magnification =\(\frac{h_2}5\) = \(\frac{120}{-60}\)

⇒ \(\frac{h_2}5\) = -2

⇒ h_{2} = -2 × 5= -10 cm

Height of image is 10 cm.

Negative sign means image is real and inverted.