Question

Два тела бросили с обрыва с одинаковой скоростью. Первое вертикально вверх, второе - вниз. Время между приземлениями этих тел 5с. Найти начальную скорость и высоту подъёма первого тела от точки падения. Высота обрыва 100м.

Answer 1

Уравнения движения для первого и второго тела:
$h_1=H_0+V_0t- \frac{1}{2} gt^2, h_2=H_0-V_0t- \frac{1}{2} gt^2$
$V_1=V_0-gt, V_2=-V_0-gt$
С момента, когда первое тело достигло максимальной высоты подъема и снова достигло высоты H_0, время его падения равно времени падения второго тела, так как к этому моменту его скорость также будет равна V_0 и направлена вертикально вниз (из закона сохранения энергии).
Значит, разницу во времени между приземлениями первое тело потратило на подъем до максимальной высоты подъема и спуск до начальной высоты H0. Его скорость при этом изменилась с V_0 на -V_0.
Тогда 
$-V_0=V_0-gT, V_0= \frac{1}{2} gT=\frac{1}{2}*10*5=25$ м/с
На достижение максимальной высоты подъема - H_1 - первое тело потратило половину от времени T - T/2 (так как время подъема с высоты H_0 до высоты H_1 равно времени спуска с высоты H_1 до высоты H_0).

Из уравнения движения первого тела
$H_1=H_0+V_0* \frac{T}{2} - \frac{1}{2} g(\frac{T}{2})^2=H_0+ \frac{gT}{2} *\frac{T}{2} - \frac{1}{8} gT^2=H_0+\frac{1}{8} gT^2=$
$100+\frac{1}{8} *10*25=131,25$м

Answer 2

Problem Analysis

We have two bodies that are thrown off a cliff with the same initial velocity. One body is thrown vertically upwards, while the other is thrown vertically downwards. The time between the bodies hitting the ground is given as 5 seconds. We need to find the initial velocity and the maximum height reached by the body thrown upwards.

Solution

Let's assume the initial velocity of both bodies is v0.

To find the initial velocity and maximum height reached by the body thrown upwards, we can use the following equations of motion:

For the body thrown upwards: - Final velocity (v) at the highest point is 0. - Acceleration (a) is equal to the acceleration due to gravity (g), which is approximately 9.8 m/s^2. - Time (t) taken to reach the highest point is half of the total time, which is 5/2 = 2.5 seconds. - Using the equation v = u + at, where u is the initial velocity, we can find the initial velocity of the body thrown upwards. - Using the equation v^2 = u^2 + 2as, where s is the maximum height reached, we can find the maximum height.

For the body thrown downwards: - The time taken to reach the ground is 5 seconds. - Using the equation s = ut + (1/2)at^2, where s is the height of the cliff, u is the initial velocity, t is the time taken, and a is the acceleration due to gravity, we can find the initial velocity of the body thrown downwards.

Let's calculate the values step by step.

Calculation

Given: - Height of the cliff (s) = 100 m - Time between the bodies hitting the ground (t) = 5 s - Acceleration due to gravity (g) = 9.8 m/s^2

Step 1: Calculate the initial velocity of the body thrown upwards (v0). - Using the equation v = u + at, where v is the final velocity (0 m/s), a is the acceleration due to gravity (-9.8 m/s^2), and t is the time taken to reach the highest point (2.5 s), we can find u. - Rearranging the equation, we have u = v - at. - Substituting the values, we get u = 0 - (-9.8 * 2.5). - Solving the equation, we find u = 24.5 m/s.

Step 2: Calculate the maximum height reached by the body thrown upwards (s). - Using the equation v^2 = u^2 + 2as, where v is the final velocity (0 m/s), u is the initial velocity (24.5 m/s), a is the acceleration due to gravity (-9.8 m/s^2), and s is the maximum height, we can find s. - Rearranging the equation, we have s = (v^2 - u^2) / (2a). - Substituting the values, we get s = (0 - (24.5)^2) / (2 * (-9.8)). - Solving the equation, we find s ≈ 30.1 m.

Step 3: Calculate the initial velocity of the body thrown downwards (v0). - Using the equation s = ut + (1/2)at^2, where s is the height of the cliff (100 m), u is the initial velocity, t is the time taken (5 s), and a is the acceleration due to gravity (-9.8 m/s^2), we can find u. - Rearranging the equation, we have u = (s - (1/2)at^2) / t. - Substituting the values, we get u = (100 - (1/2)(-9.8)(5)^2) / 5. - Solving the equation, we find u ≈ 49 m/s.

Answer

Based on the calculations, we have the following results: - The initial velocity of the body thrown upwards is approximately 24.5 m/s. - The maximum height reached by the body thrown upwards is approximately 30.1 m.

Please note that these calculations are based on the assumptions and equations mentioned above.