Question

На какой высоте скорость тела, брошенного вертикально вверх со скоростью v0, уменьшится в 5 раз?

Answer 1

Если начальная скорость v0, то конечная v0/5.По формуле перемещения при движении тела, брошенного вертикально вверх: h=(v^2 - v0^2) / -2g. подставим:
h=(v0^2/25 - v0^2) / -2g. h=24v0^2/50g. 
h=0,48*v0^2/g. (g-ускорение свободного падения).

Answer 2

Calculating the Height at Which the Speed of an Object Thrown Vertically Upward with Initial Speed v0 Will Decrease by 5 Times

To calculate the height at which the speed of an object thrown vertically upward with initial speed v0 will decrease by 5 times, we can use the following formula:

v^2 = u^2 + 2as

Where: - v = final velocity - u = initial velocity - a = acceleration due to gravity - s = displacement

First, let's find the final velocity when the speed decreases by 5 times.

The final velocity, v, when the speed decreases by 5 times will be v0/5.

Using the formula v^2 = u^2 + 2as, we can rearrange it to solve for s:

s = (v^2 - u^2) / (2a)

Substituting the values, we get:

s = (v0^2/25 - v0^2) / (2 * (-9.81))

Let's calculate the value of s using the given initial speed v0.

s = (v0^2/25 - v0^2) / (2 * (-9.81)).

This will give us the height at which the speed of the object will decrease by 5 times.