Question

С башни бросили небольшое ядро в горизонтальном направлении со скоростью 7 м/с. Определи модуль скорости и угол, который образует вектор скорости с вертикалью через 2 с. Сопротивление воздуха пренебрежимо мало, g = 10 м/с2. Ответ округлить с точностью до десятых.​

Answer 1

Ответ: 21,2 м/ 19,3°

Объяснение:Найдем горизонтальную и вертикальную составляющую скорости спустя время t=2с:

м/с - так как движение вдоль горизонтального направления равномерное, с начальной скоростью;

м/с - направлена вниз;

Полную скорость найдем как геометрическую сумму указанных составляющих:

м/с

Угол, который образует вектор скорости с вертикалью можно найти из соображений:

Откуда, искомый угол:

Answer 2

Problem Analysis

We are given the following information: - A small core is thrown horizontally from a tower with a speed of 7 m/s. - We need to determine the magnitude of the velocity vector and the angle it forms with the vertical after 2 seconds. - Air resistance can be neglected. - The acceleration due to gravity, g, is 10 m/s^2.

To solve this problem, we can use the equations of motion to find the final velocity and displacement of the core after 2 seconds. Then, we can calculate the magnitude of the velocity vector and the angle it forms with the vertical.

Solution

Let's break down the problem into steps:

Step 1: Find the horizontal displacement of the core after 2 seconds. - Since the core is thrown horizontally, its initial vertical velocity is 0 m/s. - The horizontal displacement can be calculated using the equation: displacement = initial velocity × time. - In this case, the initial velocity is 7 m/s and the time is 2 seconds. - Therefore, the horizontal displacement is: displacement = 7 m/s × 2 s = 14 m.

Step 2: Find the vertical displacement of the core after 2 seconds. - The vertical displacement can be calculated using the equation: displacement = initial velocity × time + (1/2) × acceleration × time^2. - In this case, the initial velocity is 0 m/s, the acceleration is -10 m/s^2 (negative because it acts in the opposite direction of the initial velocity), and the time is 2 seconds. - Therefore, the vertical displacement is: displacement = 0 m/s × 2 s + (1/2) × (-10 m/s^2) × (2 s)^2 = -20 m.

Step 3: Find the final velocity of the core after 2 seconds. - The final velocity can be calculated using the equation: final velocity = initial velocity + acceleration × time. - In this case, the initial velocity is 0 m/s, the acceleration is -10 m/s^2, and the time is 2 seconds. - Therefore, the final velocity is: final velocity = 0 m/s + (-10 m/s^2) × 2 s = -20 m/s.

Step 4: Calculate the magnitude of the velocity vector. - The magnitude of the velocity vector can be calculated using the Pythagorean theorem: magnitude = sqrt(horizontal velocity^2 + vertical velocity^2). - In this case, the horizontal velocity is 7 m/s and the vertical velocity is -20 m/s. - Therefore, the magnitude of the velocity vector is: magnitude = sqrt(7 m/s)^2 + (-20 m/s)^2) = sqrt(49 m^2/s^2 + 400 m^2/s^2) = sqrt(449 m^2/s^2) ≈ 21.2 m/s.

Step 5: Calculate the angle between the velocity vector and the vertical. - The angle can be calculated using the inverse tangent function: angle = arctan(vertical velocity / horizontal velocity). - In this case, the vertical velocity is -20 m/s and the horizontal velocity is 7 m/s. - Therefore, the angle between the velocity vector and the vertical is: angle = arctan(-20 m/s / 7 m/s) ≈ -70.5 degrees.

Answer

The magnitude of the velocity vector is approximately 21.2 m/s and the angle it forms with the vertical is approximately -70.5 degrees.

Note: The negative sign indicates that the angle is measured below the horizontal.