What is the maximum falling velocity of a spherical object in the atmosphere

Universal gravitation. Not only does the Earth have an attraction to the objects around it, but there is also such an attraction between any two objects. This attraction between objects is prevalent between all things in the universe and is called gravitational force.

Gravitation is an interaction between objects due to the mass they have. Its size is related to the mass of the object and the distance between the two objects. The greater the mass of the object, the greater the gravitational pull between them; The farther apart the objects, the less gravitational pull there is between them.

The gravitational force between two normally objects is so small that we don't perceive it and can ignore it. For example, two people with a mass of 60 kilograms are separated by meters, and the gravitational force between them is less than one millionth of a newton.

And an ant drags the fine grass.

The force of the terrier is 1,000 times that of this gravitational force!

However, in the celestial system, gravity plays a decisive role due to the mass and size of the celestial bodies. The Earth, which is still relatively small in the mass of celestial bodies, already has a huge impact on the gravitational pull of other objects, it binds humans, the atmosphere and all terrestrial objects to the Earth, it makes the Moon and artificial Earth satellites.

Rotate around the earth without leaving.

Gravity is caused by the gravitational pull of the earth on objects near the ground.

The force of attraction between any two objects or two particles in relation to their mass product. The most prevalent force in nature. Gravity is abbreviated and sometimes gravitational force is also weighed. In particle physics, it is called gravitational interaction and strong and weak forces.

Electromagnetic force. Collectively referred to as the 4 basic interactions. Gravitational force is the weakest of these, and the gravitational force between the two protons is only 1 1035 of the electromagnetic force between them

The gravitational pull of the proton on Earth is only 1 1010 of its electromagnetic force in a weak electric field of 1000 volts. Hence the study of interparticle interactions or particles.

in the electron microscope.

and accelerators in motion, the role of gravity is not taken into account.

The gravitational force between objects is also very small, for example, two objects with a diameter.

1 meter iron ball.

When close together.

The gravitational force is also only Newton, which is equivalent to the weight of a small drop of water in a gram.

But the mass of the Earth is large, and these two iron balls are subjected to the gravitational pull of the Earth at a rate of 4 104 Newtons, respectively.

So study the gravitational field of objects on Earth.

The gravitational pull of other surrounding objects is usually not taken into account.

There is a limit velocity, but it is related to density.

For example, if two objects of the same volume and different densities are released at the same time, when the ball reaches equilibrium, the speed of the two balls is the same and the resistance is the same, but the gravity of the large ball exceeds that of the small ball, that is, the resistance will continue to accelerate at this time.

The object is thrown vertically, the initial velocity v0=10m s, the height h=15m, and the kinetic energy theorem (v 2-v0 2) m 2=mgh, v 2-10 2) 2=10*15, v = 20m s is solved

The object first moves upwards at a speed of 10m s until the velocity drops to zero, and the formula is changed by the formula v 2 = 2 gh, 10 2 = 2 10h, h = 5m

Then do a free fall motion downward.

v'^2=2g(h+h')

v'^2=2×10×（15+5）

v'=20m/s

Therefore, the object should land at a speed of 20m s

Take up as the positive direction.

The velocity of the object at 15m is 10m s, and the time required for the velocity to change to 0 is 10-gt=0 to obtain t=1s;

And in this second, the object rises at a height of s0 = vt-1 2gt = 5m;

So in the end, the object will fall down with an initial velocity of 0 at s1 = s0 + 15 = 20m.

s1=1 2gt gives t=2s

The final velocity v=-gt=-20m s (downward).

From the square of h=1 2at, we get t=root number 3 seconds.

v=v0+at=

The speed decreases first and then increases. That is, first ascending, then falling.

v=20m|s

Let the height of the object from the ground when the suspension line is broken, and the total time of the object movement after the line is broken is t.

After the line is broken, the object moves vertically upward, and takes the vertical upward as the positive direction.

v0＝20 m/s

then h v0*t (g*t 2 2) (the "displacement" in the displacement formula is relative to the throw point).

i.e. h 20*t (10*t 2 2)Equation 1

According to the condition of the question, t 2 seconds, the displacement in the last 2 seconds is s 100 meters (because the positive direction is vertical upward).

By the vertical upward throwing motion displacement formula, known.

At the time of throwing (t 2 seconds), the displacement of the object to the thrown point is s1 v0*(t 2) [g*(t 2) 2 2 ].

i.e. s1 20*(t 2) [10*(t 2) 2 2 ] s1 is directional).

Visible, s(h) s1

100 ( h) {20*(t 2) [10*(t 2) 2 2 ]} equation 2

Combining equations 1 and 2, we can solve h 160 meters and t 8 seconds.

i.e. the height of the object at the time of the breakage of the dangling line is 160 meters, and the elapsed time of the falling of the object is 8 seconds.

Note: The positive or negative of the displacement of the thrown point is equivalent to the positive or negative of the point on the number axis, the coordinates at the throwing point are 0, the coordinates above it are positive, and the coordinates below it are negative.

This question should ignore air resistance.

When the balloon rises to a certain height at a speed of 20m s and the suspension line is disconnected, the object continues to decelerate evenly and rises to the highest point, and the rising height at this time is h1

Yes: H1 = V 2G = 20M

When the object is at the highest point, the height from the ground is h, and the time to fall to the ground is t: h=gt 2=5t

Then the height of t-2 seconds is h2=g(t-2) 2=5t -20t+20 has: h-h2=100

That is, 5t - (5t -20t + 20) = 100, get t = 6s so h = 180m

The height of the object at the time of the breakage is δh=h-h1=160mt=6s

There is a reed slow balloon at a speed of 5m s from the ground to bend the slag speed, after 30s, the height h=vt=5x30=150m

Drop a heavy object, ask for . .How long does it take for the object to be buried before it falls to the ground? The object is thrown upright upright.

h=vot-(1/2)g t ^2

150=5t-(1/2x10) t ^2

150=5t-5 t ^2

t ^2-t-30=0

t=6s

Solution: v=10m s, g=10m s 2, according to the title, there is: v=gt1, it can be seen that t1=v g=(10m s) (10m s 2)=1s

s1=gt1 2 2=10m s x (1s) 2=5mSo, s=175m+5m=180m

Yes: s=gt 2 2 obtains: t 2=2s g=2x180m 10m s, that is, t = 6s

t total = 6s + 1s = 7s

A: It takes 7 seconds to reach the ground.

Velocity = 10m s, acceleration -10m s 2, displacement is -175 m, substituted by the formula s = vt + 1 2gt 2

(1) Obtained from v0 g *t.

The rise time is obtained by v0 g 10 10 1 sec (2) at the end of v0 g*t2 at the end of t.

The velocity at the end of 5 seconds is v at the end of 10 10*5 40 m s, and the minus sign indicates that the object is falling.

3) When the object just leaves the balloon, it has a vertical upward velocity v0 10 m s, so the object is doing a vertical upward throwing motion.

From the displacement formula, s v0* t (g* t 2 2 ) (h ) 10*6 (10*6 2 2) is obtained to obtain the height of h 120 meters.

Note: The displacement is negative.

v=at

First, it decelerates uniformly to 0m s

v=10 t=1s

Then it accelerates evenly to the ground, and it takes 5 seconds to find the height of the fall t=5s

x=1/2at^2

x=125m brought in

Is that so?

You don't give the question, you see if the problem is the same.

The mechanical energy I use is conserved.

1/2)mv0^2+mgh=(1/2)mvt^2100+2400=vt^2

vt=50m/s

Finding velocity can also be done using kinematic formulas. That is, the initial velocity is 10m s, and the vertical throw is 0m s, and then the free fall is done. (vt 2-0=2gx x is the distance from the highest point to the ground, this distance is the height of 120m in the problem + the displacement of the ascending stage you want to find by yourself.)

g= can also be found, but it is not as convenient as mechanical conservation. ）

Rise to the highest point of time.

0=10-10t1

t1 = 1s descent time 50 = 10t2

t2=5st=t1+t2=6s

When an object falls from a balloon, the velocity is 10m s, and it is accelerated by the downward gravity and decelerates evenly.

Set up to be the positive direction. Yes.

s = vt-gt 2 2, where s = -120m, v = 10m s

Solve the equation and get.

t=...s ,vt=v-gt=...