Question

ПОМОГИТЕ!!!СРОЧНОО!!!Найдите скорость входа в воду прыгуна с пятиметрового трамплина, если начальная скорость отталкивания спортсмена v0 = 5 м/с

Answer 1

H=5м v0=5м/с     

v²-v0²=2*g*h
v²=v0²+2*g*h
v=√(v0²+2*g*h)
v=√(25+20*5)=11.1

Answer 2

I can help you with your physics question.

The speed of entry into the water of a diver from a five-meter diving board, if the initial speed of the athlete's push-off is v0 = 5 m/s, can be found by using the conservation of energy principle. The diver's initial kinetic energy is equal to the sum of his final kinetic energy and potential energy when he reaches the water surface. Assuming that there is no air resistance and the water level is zero, we can write:

$$\frac{1}{2}mv_0^2 = \frac{1}{2}mv^2 + mgh$$

where m is the mass of the diver, v is the speed of entry, g is the acceleration due to gravity, and h is the height of the diving board. Solving for v, we get:

$$v = \sqrt{v_0^2 + 2gh}$$

Plugging in the given values, we get:

$$v = \sqrt{(5\text{ m/s})^2 + 2(9.8\text{ m/s}^2)(5\text{ m})}$$

$$v \approx 11.1\text{ m/s}$$

Therefore, the speed of entry into the water of the diver is about 11.1 m/s.

You can also find similar problems and solutions on the web, such as [this one](http://physmatica.ru/reshebniki/reshebnik-fizika-10-klass-v-a-kasyanov/3-najdite-skorost-vxoda-v-vodu-pryguna-s-pyatimetrovogo-tramplina-esli-nachalnaya-skorost-ottalkivaniya-sportsmena-v0-5-m-s-11-1-m-s.html), [this one](https://online-otvet.ru/fizika/5cea7bf896f4e19a290ac2ba), or [this one](http://physmatica.ru/reshebniki/reshebnik-fizika-10-klass-v-a-kasyanov/2-pruzhinnyj-pistolet-vybrasyvaet-rezinovuyu-pulyu-na-vysotu-18m-nad-poverxnostyu-zemli-na-kakuyu-maksimalnuyu-vysotu-podnyalas-by-pulya-esli-by-vystrel-byl-sdelan-s-poverxnosti-marsa-47-7-m.html). I hope this helps you understand the concept. Have a nice day!