A man is moving in a hall whose surface is a reflecting surface. There is a lamp at height of 4.2 m in hall and a man while moving is looking at its spot on the floor. If the spot appears to him moving at constant speed of 2.5 m/s relative to ground velocity of man in m/s is. [Height of man is 1.6 m] (Write upto two digits after the decimal point.)

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Solution

$t a n θ = \frac{x_{1}}{4.2} = \frac{(x_{2} - x_{1})}{1.6}$

$1.6 x_{1} = 4.2 x_{2} - 4.2 x_{1}$

$\frac{5.8}{4.2} x_{1} = x_{2}$

or $v_{2} = \frac{5.8}{4.2} v_{1} = \frac{5.8}{4.2} \times 2.5$

$v_{2} = 3.45$

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tan θ=x14.2=(x2−x1)1.6 1.6x1=4.2x2−4.2x1 5.84.2x1=x2 or v2=5.84.2v1=5.84.2×2.5 v2=3.45

$t a n θ = \frac{x_{1}}{4.2} = \frac{(x_{2} - x_{1})}{1.6}$

$1.6 x_{1} = 4.2 x_{2} - 4.2 x_{1}$

$\frac{5.8}{4.2} x_{1} = x_{2}$

or $v_{2} = \frac{5.8}{4.2} v_{1} = \frac{5.8}{4.2} \times 2.5$

$v_{2} = 3.45$