A person of height 1.6 m is walking away from a lamp post of height 4 m along a straight path on the flat ground. The lamp post and the person are always perpendicular to the ground. If the speed of the person is 60 cm s$$^{-1}$$, the speed of the tip of the person's shadow on the ground with respect to the person is ______ cm s$$^{-1}$$.

Let the lamp post be at point $$O$$. The person is at point $$P$$, a distance $$y$$ (in cm) from $$O$$. The tip of the shadow is at point $$T$$, so $$PT = x$$ and $$OT = y + x$$.

Heights (in cm): lamp post = $$400$$, person = $$160$$.
Because $$\triangle OTP$$ and $$\triangle PT(\text{top of person})$$ are similar,

$$\frac{400}{\,y + x\,} = \frac{160}{\,x\,}$$.

Cross-multiplying:
$$400x = 160(y + x)$$
$$400x = 160y + 160x$$
$$240x = 160y$$
$$\Rightarrow\; x = \frac{2}{3}\,y$$ $$-(1)$$

Differentiating $$(1)$$ with respect to time $$t$$,

$$\frac{dx}{dt} = \frac{2}{3}\,\frac{dy}{dt}$$ $$-(2)$$

The person walks away at $$\frac{dy}{dt} = 60\text{ cm s}^{-1}$$. Substituting in $$(2)$$:

$$\frac{dx}{dt} = \frac{2}{3}\times 60 = 40\text{ cm s}^{-1}$$.

Speed of the tip of the shadow relative to the ground:
$$\frac{d}{dt}(y + x) = \frac{dy}{dt} + \frac{dx}{dt} = 60 + 40 = 100\text{ cm s}^{-1}$$.

Required speed (tip with respect to the person) = ground speed of tip − ground speed of person:

$$100 - 60 = 40\text{ cm s}^{-1}$$.

Hence, the speed of the tip of the shadow relative to the person is 40 cm s$$^{-1}$$.