Question

a fish F is in a pond at a depth of 0.8m from the water surface and is moving vertically upward with 2m/s. at the same instant, a bird B is aat a height of 6m from the water surface and is moving down with 3m/s. at this instant both are on the samr vertical line as shown. find height of B wrt F, depth of F wrt B, velo of B wrtF and velo of F wrt B

Answer 1

• Height of B with respect to F: $8.8\text{ m}$
• Depth of F with respect to B: $6.6\text{ m}$
• Velocity of B with respect to F: $5\text{ m/s}$
• Velocity of F with respect to B: $4.5\text{ m/s}$

Answer 2

Solution:

1. Height of B with respect to F:

   • The real height of B from the water surface is $h = 6\text{ m}$.
   • For underwater observation, the apparent height is given by $$h' = \left(\frac{\mu}{\mu'}\right)h \quad \text{with } \mu= \frac{4}{3}\text{ (water)},\; \mu'=1\text{ (air)}$$ $$h' = \frac{4}{3}\times 6 = 8\text{ m}$$
   • The fish is at a real depth $d = 0.8\text{ m}$.
   • So the height of B with respect to F (i.e. distance from F to the apparent position of B) becomes: $$8 + 0.8 = 8.8\text{ m}$$

2. Depth of F with respect to B:

   • The real depth of F is $d = 0.8$ m.
   • The apparent depth when viewed from air is given by $$d' = d\left(\frac{\mu'}{\mu}\right) = 0.8\times\frac{1}{\frac{4}{3}} = 0.8\times\frac{3}{4} = 0.6\text{ m}$$
   • Adding the bird’s real height gives the depth of F with respect to B: $$6 + 0.6 = 6.6\text{ m}$$

3. Velocity of B with respect to F:

   • Fish’s real upward speed: $v_F = 2\text{ m/s}$.
   • Bird’s real downward speed: $v_B = 3\text{ m/s}$.
   • When seen by F, the bird’s velocity is not altered by refraction (since, while image position scales, the time derivatives add directly for the underwater observer). Thus, $$v_{B\,\text{relative to }F} = v_F + v_B = 2 + 3 = 5\text{ m/s}$$

4. Velocity of F with respect to B:

   • For the bird, the apparent velocity of the fish is reduced by the factor $\frac{\mu'}{\mu}$: $$\text{Apparent }v_F = 2\times\frac{1}{\frac{4}{3}} = 2\times\frac{3}{4} = 1.5\text{ m/s}$$
   • Thus, the relative velocity as observed by B is: $$v_{F\,\text{relative to }B} = \text{Apparent }v_F + v_B = 1.5 + 3 = 4.5\text{ m/s}$$

Minimal Explanation:

• Compute apparent height of B in water:
  $h' = \frac{4}{3}\times6 = 8\text{ m}$; then add fish depth $0.8\text{ m}$ to get $8.8\text{ m}$.
• Compute apparent depth of F in air:
  $d' = 0.8\times\frac{3}{4} = 0.6\text{ m}$; then add bird height $6\text{ m}$ to get $6.6\text{ m}$.
• Relative velocities: For F, add the speeds directly: $2+3=5\text{ m/s}$. For B, scale $v_F$ by $\frac{3}{4}$ and add $v_B$: $1.5+3=4.5\text{ m/s}$.