A man walking with a speed of 3 m/s passes under a lamp hanging at a height of 8 m above the ground. If the height of the man is 2 m, the velocity in m/s with which the edge of the shadow of the man's head moves over the ground is

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Solution

Let H be the height of the lamp and h be the height of the man

Then, from the figure,

$\frac{H}{x_{s}} = \frac{h}{x_{s} - x} = tan θ \Rightarrow H (x_{s} - x) = h x_{s} \Rightarrow x_{s} (H - h) = H x \Rightarrow x_{s} = \frac{H}{H - h} x T a k i n g d e r i v a t i v e o n b o t h s i d e s u_{s} = \frac{H}{H - h} u = \frac{8}{8 - 2} \times 3 m / s \Rightarrow u_{s} = 4 m / s$

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A man walking with a speed of 3 m/s passes under a lamp hanging at a height of 8 m above the ground. If the height of the man is 2 m, the velocity in m/s with which the edge of the shadow of the man's head moves over the ground is

Let H be the height of the lamp and h be the height of the man Then, from the figure, Hxs = hxs − x = tan θ⇒H (xs − x) = hxs⇒xs (H − h) = Hx⇒xs = HH − hxTaking derivative on both sidesus = HH − hu = 88 − 2×3 m/s⇒us = 4 m/s