How much time does it take for a baby elephant to land in free fall

The fall of a 1000kg elephant is indeed a free fall, falling from a place of 125m, from the free fall formula 1 2gt 2=h to calculate the time is exactly 5s, the ants fall from the same place It takes a long time because the resistance of the air is taken into account, the meaning of the question is that the elephant should ignore the air resistance, so the time should also be 5s, otherwise this problem is not easy to do (the question does not give a functional relationship between air resistance and velocity).

The resistance f of the air is directly proportional to the surface area of the object f - -s

The mass of the object m is proportional to s s m 2 and s 3

So f--m 2 to the 3rd power. f=km 2 3 k is a constant.

Acceleration a=g-f m

When m=1000kg, a=g-(km 2 3) m, a=2h t 2=2*125, 5*5=10m s 2;

a==g-km^-(1/3)=g-k1000^-(1/3)=g-k/10=10 (1)

Normal g=10ms2 so k10<= k<=

When m=10g=a=g-(km 2 3) m,a=2h t 2=2*125 10*10=;

a==g-km^-(1/3)= (2)

When m=20g=, a=g-(km 2 3) m, a==g-km -(1 3)=

t = under the root number (2h a) = under the root number (2*125.)

When m=500kg, a=g-(km2, 3) m, a==g-km-(1 3)=g-k500 -(1 3)=g-[under the root number, (500)=

t = under the root number (2h a) = under the root number (2*125.)

20g. 500kg.

You don't have to look at it, it's 5s

The ant uses 10s because the resistance of the air is very strong relative to the gravity on it, which has a great impact on its speed.

And the 1000kg, 500kg elephant is negligible in terms of air resistance relative to body weight, so the time is the same.

5 seconds. The time taken for a free fall is only the distance of the object from the ground

Formula: h=1 2*g*t*t, this formula is applied to objects with relatively large masses and relatively small volumes (i.e., ignoring air resistance). The fall of the ants cannot ignore the air resistance, so the calculation will be very complicated.

Be careful. In our daily lives, there is air resistance. Don't forget, landlord.

And the topic is not complete. ~~

Solution: Set the drop height to h

Because 2GH=VT 2

So h=vt 2 2g

Because H2=(1 2)GT 2

So (vt 2 2g) 2=(1 2)gt 2t=vt (root number 2)g

Try to look at it,

Cotton experiences more air resistance than iron.

In fact, the heavier ones hit the ground first than the lighter ones. I don't think that theory is quite right, 2 are the same size, iron seeking.

and plastic, but the iron one fell to the ground first, and I got there N times

In principle, according to the book, the force of free fall and downward is the same.

TM's you don't know there's resistance. Does cotton have the same resistance to air as iron?

Are you troubled by this? Why don't you bother sending here?

2gs=v2, with a value of 2 times 10 times 45 = speed squared, velocity = gt, ok?

About 1 minute! But it's best to experience it for yourself!

When the time is set to be free fall, the relationship between the falling height and the time is h=1 2gt

Then t=sqrt(2h g)=under the root number.

by h=1 2gt

t= (2h g) [is the root number].

Approximately equal.

I'm telling you when you jump off it.

At the beginning, the height of the object from the ground is h

Let : The landing time is 2t, and half of the landing time is t by h = get:

When the exercise time is half of the landing time. Drop distance.

h1 = distance from the starting point to the ground.

h = 4 【

So h1 = 1 4)h

A uniform acceleration motion with an initial velocity of 0.

The first t, the second t, the third t... The displacement ratio within the nth t is 1:3:5:. odd ratio).

The first half of the time is t (the first t), the second half of the time is also t (the second t), and the displacement ratio is 1:3

So the displacement in the first half of the time compared to the displacement of the whole segment.

That's 1 : 3 + 1 = 1 : 4

Therefore, when the movement time is half of the landing time. The drop distance is a quarter of h.

The distance of the fall is 1 4h

Let the vertical height be h, then the total time taken to fall is (2h g) 1 2, and when the movement time is half of the total falling time, the time taken is 1 2 (2h g) 1 2, combined with the formula h'=1 2gt 2, get h'=1/4h.

Free fall time = root number 2gs

Air resistance is not counted:

t = change sign (2h g).

That's about speed, it's GT, it's 316m s

s=1 2gt 2 Push out t=10 times the root number for 10 seconds.

v = gt = 10 * 10 times the root number 10 = 100 times the root number 10 m s

First of all, it should be known that the velocity of an object in free fall has nothing to do with its mass.

For objects that do free fall motion. H = gt 2 2 and v = gt.

From both formulas, it can be seen that both height and landing speed are related to time.

It is estimated that your question is a practical problem and should be considered for air resistance. So what you're given is just a theoretical calculation answer.

The landing time is seconds, and the landing speed is 316 meters per second.

However, these are theoretical calculations, and air resistance is not taken into account, so the velocity of this object when it lands must be much less than this value, and it will not be so large. And the landing time will be more than calculated.

Set the time required for landing to t

Instantaneous velocity in the middle = average velocity.

is the middle moment in the last 1 second.

t'= of the instantaneous velocity v'= Average velocity in the last 1 second = 100m 1s = 100m s

gt'=v'=10(

t = drop height h = (1 2) gt 2 =

The speed of landing v=gt=10*

The displacement of the passage in the last 1 second is 100 meters, the average speed of the passage in the last 1 second is 100 1 = 100, and the speed of landing v = 100 + g * (1 2) = 105

Drop height hmgh = (1 2) mV

h=..gt=v

t=v/g=..

Let the distance of the object fall be h, and the time of fall is t, then h=gt2 2, and when the time of the object falling is half of the total time, the displacement is g(t2 4)=h 4

Set the height h

h=gtt/2

Half the time. h=g*(t 2)(t 2) 2=(gtt 2) 4=h 4 The fall distance is 1 4 of the original height