Question

Помогите решить (Воздушный шар равномерно поднимается со скоростью 2м/с.На высоте 7м от поверхности земли выбросили груз.Через какой промежуток времени груз упадет на землю.Какой будет его скорость в момент падения?)Если можно то розпешите по детальней

Answer 1

H=h0+v0*t+at^2/2
h=0 
h0=7м
v0=2
a=-10
0=7+2*t-5t^2
5t^2-2t-7=0
d=4+140=144=12^2
t=(2+12)/10=1,4 сек - время падения
v=v0+at
v=2-10*t
v=2-10*1,4=-12 м/с - скорость в момент падения

Answer 2

Problem Analysis

We are given that an air balloon is rising uniformly with a speed of 2 m/s. At a height of 7 m above the ground, a weight is released. We need to find the time it takes for the weight to fall to the ground and its velocity at the moment of impact.

Solution

To solve this problem, we can use the equations of motion for uniformly accelerated motion. The equations we will use are:

1. Displacement equation: `s = ut + (1/2)at^2` 2. Final velocity equation: `v = u + at` 3. Initial velocity equation: `u = 0` (since the weight is released from rest) 4. Acceleration equation: `a = g`, where `g` is the acceleration due to gravity (approximately 9.8 m/s^2)

Let's solve for the time it takes for the weight to fall to the ground first.

Finding the Time of Impact

We know that the initial velocity (`u`) is 0 m/s, the acceleration (`a`) is -9.8 m/s^2 (negative because it acts in the opposite direction to the motion), and the displacement (`s`) is 7 m. We need to find the time (`t`).

Using the displacement equation, we can rearrange it to solve for `t`:

s = ut + (1/2)at^2
7 = 0*t + (1/2)*(-9.8)*t^2
7 = -4.9t^2
t^2 = -7/(-4.9)
t^2 = 1.4286
t ≈ √1.4286
t ≈ 1.195 s

Therefore, it takes approximately 1.195 seconds for the weight to fall to the ground.

Finding the Velocity at Impact

To find the velocity of the weight at the moment of impact, we can use the final velocity equation. We know that the initial velocity (`u`) is 0 m/s, the acceleration (`a`) is -9.8 m/s^2, and the time (`t`) is approximately 1.195 s. We need to find the final velocity (`v`).

Using the final velocity equation, we can solve for `v`:

v = u + at
v = 0 + (-9.8)*1.195
v ≈ -11.71 m/s

Therefore, the velocity of the weight at the moment of impact is approximately -11.71 m/s. The negative sign indicates that the velocity is directed downwards.

Conclusion

To summarize: - The weight will take approximately 1.195 seconds to fall to the ground. - The velocity of the weight at the moment of impact will be approximately -11.71 m/s (directed downwards).

Please let me know if anything is unclear or if you need further assistance!