Answer: 0 m/s or $0.8 \mathrm{~m} / \mathrm{s}$.

Solution. The mosquito flew a distance of $s=v t=0.5 \cdot 20=10$ m before landing.

That is, it moved along the trajectory $A B$.

![](https://cdn.mathpix.com/cropped/2024_05_06_f717c0793dcd8afe888ag-11.jpg?height=268&width=580&top_left_y=2330&top_left_x=818)

The cosine of the angle between the trajectory and the water surface is 0.8.

Obviously, the speed of the shadow moving along the bottom coincides with the speed of the shadow moving on the surface of the pond.

Since the angle of incidence of the sunlight coincides with the angle between the trajectory and the water surface, there are two possible scenarios.

The first - the mosquito is moving along the sunbeam.

![](https://cdn.mathpix.com/cropped/2024_05_06_f717c0793dcd8afe888ag-12.jpg?height=277&width=559&top_left_y=615&top_left_x=834)

The speed of the shadow on the bottom of the pond $v_{\text {shadow }}=0 \mathrm{~m} / \mathrm{s}$.

The second - the mosquito is flying towards the sunbeam. 2

![](https://cdn.mathpix.com/cropped/2024_05_06_f717c0793dcd8afe888ag-12.jpg?height=233&width=644&top_left_y=1137&top_left_x=932)

The speed of the shadow $v_{\text {shadow }}=2 v \cos \beta=2 \cdot 0.5 \cdot 0.8=0.8 \mathrm{~m} / \mathrm{s}$.
8. (15 points) A mosquito was moving over the water in a straight line at a constant speed of \( v = 0.5 \) m/s and at the end of its movement, it landed on the water surface. 20 seconds before landing, it was at a height of \( h = 6 \) m above the water surface. The cosine of the angle of incidence of the sunlight on the water surface is 0.6. The incident sunlight, which creates the shadow of the mosquito, and its trajectory lie in the same vertical plane. Determine the speed at which the shadow of the mosquito moved along the bottom of the water body.