Question

Два тела брошены вертикально вверх из одной точки на поверхности земли одно вслед за другим через 0,5 с. чепез сколько секунд после броска первого тела оба иела встретятся, если их начальная скорость одинаковы и равны 10 м/с

Answer 1

Δt=0.5 c    vo=10 м/с    t=?
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y1=vo*t-g*t²/2
y2=vo*(t-Δt)-g*(t-Δt)²/2
При встрече их координаты равны
y1=y2
vo*t - g*t²/2=vo*(t-Δt) - g*(t-Δt)²/2
vo*t - g*t²/2=vo*t - vo*Δt - g*t²/2 + 2*g*t*Δt/2 - g*Δt²/2
t=vo/g + Δt/2=10/10 + 0.5/2=1.25 c
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Answer 2

Problem Analysis

We have two bodies thrown vertically upwards from the same point on the surface of the Earth, one after the other, with the same initial velocity of 10 m/s. We need to determine the time at which both bodies will meet.

Solution

To solve this problem, we can use the equations of motion for vertical motion under constant acceleration. The acceleration due to gravity near the surface of the Earth is approximately 9.8 m/s^2.

Let's consider the motion of the first body. The initial velocity is 10 m/s, and the acceleration is -9.8 m/s^2 (negative because it acts in the opposite direction to the motion). We can use the equation:

v = u + at

where: - v is the final velocity (which will be zero when the body reaches its maximum height), - u is the initial velocity, - a is the acceleration, and - t is the time.

For the first body, the final velocity is 0 m/s, the initial velocity is 10 m/s, and the acceleration is -9.8 m/s^2. We can rearrange the equation to solve for time:

t = (v - u) / a

Substituting the values, we get:

t = (0 - 10) / -9.8

Simplifying, we find:

t = 1.02 seconds

So, the first body will take approximately 1.02 seconds to reach its maximum height.

Now, let's consider the motion of the second body. The initial velocity is also 10 m/s, and the acceleration is -9.8 m/s^2. The time for the second body to reach its maximum height will be the same as the time for the first body, which is 1.02 seconds.

After reaching their maximum heights, both bodies will start falling downwards. The time taken for the second body to reach the same height as the first body will be the same as the time taken for the first body to fall from its maximum height to the ground.

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,

we can calculate the displacement of the first body from its maximum height to the ground. The initial velocity is 0 m/s (at the highest point), the acceleration is -9.8 m/s^2, and the time is 1.02 seconds. Substituting the values, we get:

s = 0 + (1/2)(-9.8)(1.02)^2

Simplifying, we find:

s = -5.02 meters

The negative sign indicates that the displacement is in the opposite direction to the initial motion.

Now, we can calculate the time taken for the second body to fall from its maximum height to the ground. The displacement is -5.02 meters (opposite to the initial motion), the initial velocity is 10 m/s, and the acceleration is -9.8 m/s^2. Using the equation:

s = ut + (1/2)at^2

we can rearrange the equation to solve for time:

t = (-u ± √(u^2 - 2as)) / a

Substituting the values, we get:

t = (-10 ± √(10^2 - 2(-9.8)(-5.02))) / -9.8

Simplifying, we find:

t ≈ 1.02 seconds

Therefore, both bodies will meet approximately 1.02 seconds after the first body is thrown.

Conclusion

Both bodies will meet approximately 1.02 seconds after the first body is thrown.