Given that hm=6ft, hl=15ft, vm=9ft/sec. We have to calculate vs (speed of shadow).
After time t secs, the man would have moved distance 9t ft away from the lamp. Let the shadow move a distance of x ft in time t secs.
Consider ∆AEC and ∆BED
∠AED=∠BED=θ (common angle)
Therefore, ∆AEC~∆BED by AA criteria
Substituting values, we get
Simplifying the equation, we get
x=6t – (1)
Differentiating (1) with respect to t, we get
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