Question

A diver $D$ is still under water $(\mu = \frac{4}{3})$ at a depth $d=10$ m. A bird is diving along line $AB$ at a constant velocity in air. When the bird is exactly above the diver he sees it at a height of 50 m from himself and velocity of the bird appears to be inclined at $45^\circ$ to the horizontal. At what distance from the diver the bird actually hits the water surface (roundoff answer to nearest integer).

Answer 1

40 m

Answer 2

Solution

1. Determine the apparent altitude:

   For a flat interface, an underwater observer sees an object in air at an apparent height

   $z_a = d + (H/\mu)$.

   At the instant when the bird is “directly above” the diver the diver sees it at a distance 50 m from him.

   Thus,

   $d + H/\mu = 50$.

   Given $d = 10$ m and $\mu = 4⁄3$,

   $H/ (4⁄3) = 50 – 10 = 40 \implies H = (4⁄3) \times 40 = 160⁄3$ m.

2. Relate the actual and apparent velocity components:

   A simple “ray–differential” treatment shows that the underwater image of a bird at $(x, H)$ is

   $A = (x, d + H/\mu)$,

   so that

   $(dx_a/dt, dz_a/dt) = (dx/dt, (1/\mu) dH/dt)$.

   We are told that the apparent velocity is inclined at $45^\circ$ to the horizontal. This gives

   $|dz_a/dt| = |dx_a/dt|$

   $\implies (1/\mu)|v_z| = v_x$

   $\implies |v_z| = \mu v_x = (4/3)v_x$.

3. Find time from the “above–diver” instant to hitting water:

   At $t = 0$ the bird is at $B(0) = (0, H)$ with $H = 160/3$ m. Its speed components are

   $v_x = V$ and $v_z = –(4⁄3)V$ (negative as it’s descending).

   The time $T$ to reach the water surface $(z = 0)$ is found from

   $H + v_z T = 0 \implies (160/3) – (4/3)V \cdot T = 0$

   $\implies T = (160/3) / ((4/3)V) = 40/V$.

4. Horizontal distance from the diver:

   During time $T$ the bird moves horizontally

   $x = V \cdot T = V \cdot (40/V) = 40$ m.

Thus, the bird actually hits the water 40 m from the diver.

Explanation (minimal):

• Use the apparent-image formula $d + H/\mu = 50$ to get $H = (4⁄3)(50–10) = 160⁄3$.
• The velocity mapping gives $|v_z| = (4⁄3)v_x$.
• Time to descend: $T = H/|v_z| = (160/3)/((4/3)V) = 40/V$, so horizontal shift = $V \cdot T = 40$ m.