Question

Тіло кинуто вертикальнр вгору зі швидкістю 16м/с. В які моменти часу тіло буде перебувати на висоті,що длрівнює половині максимальної?

Answer 1

Ответ:

вот ответ.........думаю правельно

Answer 2

Problem Analysis

We are given that an object is thrown vertically upwards with an initial velocity of 16 m/s. We need to determine the time at which the object will be at a height equal to half of its maximum height.

Solution

To solve this problem, we can use the equations of motion for an object thrown vertically upwards. The key equation we will use is:

h = u*t - (1/2)*g*t^2

where: - h is the height of the object at time t - u is the initial velocity of the object (16 m/s in this case) - g is the acceleration due to gravity (approximately 9.8 m/s^2) - t is the time

We are given that the object is at a height equal to half of its maximum height. Let's denote this height as h_max/2. We can set up the equation as follows:

h_max/2 = u*t - (1/2)*g*t^2

Now, let's solve this equation for t.

Calculation

Substituting the given values into the equation, we have:

h_max/2 = 16*t - (1/2)*9.8*t^2

Simplifying the equation, we get:

0 = 9.8*t^2 - 16*t + h_max/2

This is a quadratic equation in terms of t. We can solve it using the quadratic formula:

t = (-b ± sqrt(b^2 - 4ac)) / (2a)

where: - a = 9.8 - b = -16 - c = h_max/2

Since we are only interested in the positive value of t (time cannot be negative), we can ignore the negative solution.

Answer

The object will be at a height equal to half of its maximum height at the time given by the positive solution of the quadratic equation.