Question

Тело бросили вертикально вверх со скоростью 20 м/с. На какой высоте от первоначального уровня его скорость станет в 4 раза меньше первоначальной? Сопротивление воздуха не учитывайте.

Answer 1

По закону сохранения полной механической энергии:
Е= Ек +Еп=E к (макс)=mv²/2=200m
Ек=12,5 m
Eп= mgh
12,5m+mgh=200m
mgh=187,5m
gh=187,5
h=18,75 м

Answer 2

Problem Analysis

We are given that an object is thrown vertically upward with an initial velocity of 20 m/s. We need to find the height at which its velocity becomes 4 times smaller than the initial velocity. We are also told to ignore air resistance.

Solution

To solve this problem, we can use the equations of motion for uniformly accelerated motion. Since the object is thrown vertically upward, we can consider the acceleration due to gravity as -9.8 m/s^2.

Let's assume the initial velocity of the object is v0 = 20 m/s and the final velocity when the object reaches a certain height is vf. We need to find the height at which the final velocity becomes 4 times smaller than the initial velocity.

Using the equation for final velocity in uniformly accelerated motion:

vf = v0 + at

where: - vf is the final velocity - v0 is the initial velocity - a is the acceleration - t is the time

Since the object is thrown vertically upward, the acceleration is equal to the acceleration due to gravity, which is -9.8 m/s^2. Therefore, the equation becomes:

vf = v0 - 9.8t We want to find the height at which the final velocity becomes 4 times smaller than the initial velocity. In other words, we want to find the height at which:

vf = v0/4

Substituting the values into equation we get:

v0/4 = v0 - 9.8t

Simplifying the equation, we have:

9.8t = 3v0/4

Solving for t, we get:

t = (3v0)/(4 * 9.8)

Now, we can use the equation for displacement in uniformly accelerated motion to find the height at time t:

s = v0t + (1/2)at^2

Substituting the values, we have:

s = v0 * (3v0)/(4 * 9.8) + (1/2) * (-9.8) * ((3v0)/(4 * 9.8))^2

Simplifying the equation, we get:

s = (9v0^2)/(4 * 9.8) - (9v0^2)/(4 * 9.8 * 4)

s = (9v0^2)/(4 * 9.8) - (9v0^2)/(16 * 9.8)

s = (9v0^2)/(4 * 9.8) - (9v0^2)/(4 * 9.8)

s = 0

Therefore, the height at which the object's velocity becomes 4 times smaller than the initial velocity is 0 meters. This means that the object will reach its maximum height and start falling back down when its velocity becomes 4 times smaller than the initial velocity.

Answer

The height at which the object's velocity becomes 4 times smaller than the initial velocity is 0 meters.