Given that h_m=2m, h_l=5m, v_m=4.8km/hr=4800m/hr. We have to calculate v_s(speed of shadow).

After time t hrs, the man would have moved distance 4800t m away from the lamp. Let the shadow move a distance of x m in time t hrs.

Consider ∆AEC and ∆BED

∠AED=∠BED=θ (common angle)

∠EAC=∠EBD=90°

Therefore, ∆AEC~∆BED by AA criteria

Substituting values, we get

Simplifying the equation, we get

x=3200t – (1)

Differentiating (1) with respect to t, we get

v_s=3.2km/hr