Question

мяч отскочил от стены на 3м долетел до пола и поднялся вверх на 2м. найдите путь и перемещение мяча? как правильно высчитать ?

Answer 1

Ответ:

путь=5м

перемещение=1м

Объяснение:

h1=3м

h2=2м

Путь – это дина траектории. В данном случае путь равен

Путь=h1 + h2 =3 + 2 =5 м

Перемещение – это вектор, соединяющий начальное и конечное положение точки.

Перемещение= h1 –h2 =3 - 2 =1 м

Answer 2

Problem Statement

A ball bounces off a wall and reaches a height of 3m before falling to the ground and rebounding to a height of 2m. We need to find the path and displacement of the ball.

Solution

To calculate the path and displacement of the ball, we can break down the motion into two parts: the upward motion and the downward motion.

1. Upward Motion: - The ball reaches a height of 3m. - Let's assume the initial velocity of the ball when it leaves the ground is u m/s. - The final velocity at the highest point is 0 m/s (since the ball momentarily stops before falling back down). - The acceleration due to gravity is -9.8 m/s^2 (negative because it acts in the opposite direction to the motion). - Using the kinematic equation: v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement, we can solve for u. - Plugging in the values, we get: 0^2 = u^2 + 2(-9.8)(3). - Solving for u, we find: u = sqrt(2 * 9.8 * 3) m/s. - The time taken to reach the highest point can be calculated using the equation: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. - Plugging in the values, we get: 0 = sqrt(2 * 9.8 * 3) + (-9.8)t. - Solving for t, we find: t = sqrt(2 * 3 / 9.8) s. - The displacement during the upward motion is equal to the height reached, which is 3m.

2. Downward Motion: - The ball falls from a height of 3m to a height of 2m. - The initial velocity when the ball starts falling is 0 m/s (since it momentarily stops at the highest point). - The final velocity when the ball reaches a height of 2m can be calculated using the equation: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. - Plugging in the values, we get: v = 0 + (-9.8)t. - The time taken to fall from 3m to 2m can be calculated using the equation: s = ut + (1/2)at^2, where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time. - Plugging in the values, we get: 2 = 0 * t + (1/2)(-9.8)t^2. - Solving for t, we find: t = sqrt(2 / 9.8) s. - The displacement during the downward motion is equal to the difference in height, which is 3m - 2m = 1m.

Path and Displacement of the Ball

The path of the ball can be visualized as follows: - The ball starts at the ground (0m). - It reaches a height of 3m and momentarily stops. - It falls back down to a height of 2m.

The displacement of the ball is the difference between the initial and final positions. In this case, the ball starts and ends at the ground, so the displacement is 0m.

To summarize: - The path of the ball is from the ground to a height of 3m, and then back down to a height of 2m. - The displacement of the ball is 0m.

Please note that the calculations provided above are based on the assumption of idealized motion, neglecting factors such as air resistance.