Question

Напишите плс решение. С вышки брошен камень в горизонтально направлении со скоростью 20 м/с. Через сколько секунд скорость камня увеличится в два раза? Какой угол с вертикалью будет составлять вектор скорости в этот момент?

Answer 1

Vx=20 м/с     t=?
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v=2*vx  горизонтальная скорость
vy=g*t   вертикальная скорость
v²=(2*vx)²=vx²+(g*t)²
t=√(4vx²-vx²)/g=vx*√3/g=20*√3/2≈3.46 c
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Answer 2

Problem Analysis

We are given that a stone is thrown horizontally from a tower with a velocity of 20 m/s. We need to determine the time it takes for the stone's velocity to double and the angle the velocity vector makes with the vertical direction at that moment.

Solution

To solve this problem, we can use the equations of motion for projectile motion. Since the stone is thrown horizontally, the initial vertical velocity is zero.

Let's denote: - v as the velocity of the stone, - v0 as the initial velocity of the stone, - g as the acceleration due to gravity (approximately 9.8 m/s^2), - t as the time taken, - θ as the angle with the vertical direction.

Using the equation for horizontal motion, we have: v = v0

Using the equation for vertical motion, we have: v = gt

Since the stone's velocity doubles, we can write: 2v0 = gt

Simplifying the equation, we find: t = 2v0 / g

To find the angle θ, we can use the equation: tan(θ) = v / v0

Substituting the values, we have: tan(θ) = gt / v0

Simplifying the equation, we find: θ = arctan(gt / v0)

Now, let's calculate the values.

Calculation

Given: - Initial velocity v0 = 20 m/s - Acceleration due to gravity g = 9.8 m/s^2

Using the equation t = 2v0 / g, we can calculate the time: t = 2 * 20 / 9.8 ≈ 4.08 seconds

Using the equation θ = arctan(gt / v0), we can calculate the angle: θ = arctan(9.8 * 4.08 / 20) ≈ 1.00 radians

Answer

The velocity of the stone will double after approximately 4.08 seconds. At that moment, the angle the velocity vector makes with the vertical direction will be approximately 1.00 radians.

Please note that the calculations are approximate and may vary slightly depending on the exact values used for acceleration due to gravity and the initial velocity of the stone.

Let me know if you need any further assistance!