Question

ПОЖАЛУЙСТА ПОМОГИТЕ Уравнения координат груза, сброшенного с горизонтального летящего самолета, имеют вид: y = 120 – 2t2 и x = 300t, где x и y – координаты груза, время дано в секундах. С какой высоты сброшено тело? Как далеко упадет оно от места сброса? (Сопротивлением воздуха пренебречь). Ответ округлить до целого числа. Высота: м Дальность полета: м

Answer 1

Ответ:

Высота:  

120 м

Дальность полета:  

1470 м

Объяснение:

Answer 2

Problem Analysis

We are given the equations for the coordinates of a cargo dropped from a flying plane: - y = 120 - 2t^2 - x = 300t

We need to find the height from which the cargo was dropped and the distance it will fall from the point of release. We can solve this problem using basic kinematic equations.

Solution

To find the height from which the cargo was dropped, we need to find the maximum value of y. The equation y = 120 - 2t^2 represents a parabolic path, and the maximum height occurs at the vertex of the parabola.

The vertex of a parabola with equation y = ax^2 + bx + c is given by the formula: t = -b / (2a)

In our case, a = -2, b = 0, and c = 120. Substituting these values into the formula, we can find the time at which the cargo reaches its maximum height.

To find the distance the cargo will fall from the point of release, we need to find the time it takes for the cargo to reach the ground. This can be done by setting y = 0 and solving for t.

Let's calculate the height and distance using the given equations.

Calculation

1. Finding the height: - The equation for the height is y = 120 - 2t^2. - The vertex of the parabola is given by t = -b / (2a). - Substituting the values, we get t = -0 / (2 * -2) = 0. - The maximum height occurs at t = 0. - Substituting t = 0 into the equation y = 120 - 2t^2, we get y = 120 - 2(0)^2 = 120. - Therefore, the height from which the cargo was dropped is 120 meters.

2. Finding the distance: - The equation for the distance is x = 300t. - To find the time it takes for the cargo to reach the ground, we set y = 0. - Substituting y = 0 into the equation y = 120 - 2t^2, we get 0 = 120 - 2t^2. - Solving this equation, we get t = ±√(120/2) = ±√60 ≈ ±7.746. - Since time cannot be negative in this context, we take t = 7.746. - Substituting t = 7.746 into the equation x = 300t, we get x = 300 * 7.746 ≈ 2323.8. - Therefore, the distance the cargo will fall from the point of release is 2324 meters.

Answer

The cargo was dropped from a height of approximately 120 meters. It will fall a distance of approximately 2324 meters from the point of release.

Please note that these values are approximate and rounded to the nearest whole number.