Solution
The “flare” is the image of the lamp formed by reflection on the water surface. The position of the reflection point is determined by the geometry of the incident and reflected rays, which form similar triangles with the vertical heights of the lamp and the observer.
1. Relation between positions:
Let the lamp be at $x=0$.
Let the flare be at distance $x_f$ from the lamp.
Let the boy be at distance $x_b$ from the lamp.
From similar triangles: $$ \frac{H}{x_f} = \frac{h}{x_b – x_f} $$ $$ H(x_b – x_f) = h x_f $$ $$ H x_b = x_f(H + h) \implies x_f = x_b \left( \frac{H}{H+h} \right) $$
Let the lamp be at $x=0$.
Let the flare be at distance $x_f$ from the lamp.
Let the boy be at distance $x_b$ from the lamp.
From similar triangles: $$ \frac{H}{x_f} = \frac{h}{x_b – x_f} $$ $$ H(x_b – x_f) = h x_f $$ $$ H x_b = x_f(H + h) \implies x_f = x_b \left( \frac{H}{H+h} \right) $$
2. Relation between velocities:
Differentiating with respect to time ($v = dx/dt$): $$ v_{flare} = v_{boy} \left( \frac{H}{H+h} \right) $$
Differentiating with respect to time ($v = dx/dt$): $$ v_{flare} = v_{boy} \left( \frac{H}{H+h} \right) $$
3. Calculation:
Given $H=5.4$ m, $h=1.8$ m, $v_{flare} = 2.4$ m/s.
Ratio term: $$ \frac{H}{H+h} = \frac{5.4}{5.4 + 1.8} = \frac{5.4}{7.2} = \frac{3}{4} $$ Substitute values: $$ 2.4 = v_{boy} \times \frac{3}{4} $$ $$ v_{boy} = \frac{2.4 \times 4}{3} = 3.2 \text{ m/s} $$
Given $H=5.4$ m, $h=1.8$ m, $v_{flare} = 2.4$ m/s.
Ratio term: $$ \frac{H}{H+h} = \frac{5.4}{5.4 + 1.8} = \frac{5.4}{7.2} = \frac{3}{4} $$ Substitute values: $$ 2.4 = v_{boy} \times \frac{3}{4} $$ $$ v_{boy} = \frac{2.4 \times 4}{3} = 3.2 \text{ m/s} $$
Answer: 3.2 m/s