Senior One Physics, Free Fall Motion, Objects Do Free Fall Motion Senior One Physics

The speed of the T2 wooden stick on the top of the 3rd floor is V1, the speed at the bottom of the 3rd floor is V2, the speed at the bottom of the 2nd floor is V3, and the free fall is an acceleration motion When V1 passes through the 3rd floor, the speed is accelerated from V1 to V2, and when it passes through the 2nd floor, the speed is accelerated from V2 to V3, so the average speed of the 3rd floor is lower than that of the 2nd floor. The height of each floor is the same, so it is faster to cross the 2nd floor, so t2

t1>t2

The best way is the image method, considering the air resistance, the wooden stick from the fifth floor approximate to do a uniform acceleration movement, just draw a V-T sketch;

Or, let the initial velocity of the third floor be v1 and the end velocity of the third floor be v2; The initial speed of the second floor is V3, the speed of the end of the second floor is V4, and the height of the first floor is H.

Judging by the average velocity, the displacement of the passing building is equal.

Descend to the top edge of the third floor at V1

Then fall between the 2nd and 3rd floors at v2

Then go down to the bottom edge of the 2nd floor at v3

v1＜v2＜v3

Average speed (v1+v2) after the third floor 2

The average speed of passing through the second floor (v2+v3) is 2

It's clear that the average speed on the third floor is the average speed on the second floor.

So t1 t2

t1〉t2

Free fall motion is a motion with uniform acceleration. Then the average speed passing through the third floor is less than the average speed passing through the second floor. The height of the building is the same, so t1 t2

t1〉t2

Passing through the third floor is the same as passing through the second floor, the distance is the same, the acceleration is the same, the initial velocity is greater on the second floor than on the third floor, so t1 t2

Acceleration problem, each floor is the same height. Free fall is a uniform acceleration motion with an acceleration g. The speed increases evenly. t=s÷v

The same height depends on the speed. The speed is large and the time is short.

Directly push the proportional formula with you, and then use this proportional formula to solve the problem (because the person who made the question is actually nothing more than to examine your proportional formula).

Derivation: Let the total displacement be divided into n segments, and the displacement of each segment is s, and the first segment: 1 2 gt 1=s

The first two paragraphs: 1 2gt 2=2s

The first three segments: 1 2gt 3 = 3s

First n segments: 1 2gt n = ns

After n-1 formulas, compared with the first formula, the total time taken for the first n segments can be obtained, so the time spent on each segment is tn-tn-1

This gives the ratio of the same displacement time as 1: 2-1: 3- 2:n- (n-1).

Then you use this proportional relationship to solve this kind of problem, and you can pass the test, and you can directly use this proportional relationship:

Solution: The ratio of time by the same displacement is 1: 2-1: 3- 2:n- (n-1) this.

Proportional Relationship: Available: ......

Thus ......That is, what is sought.

Answer: slightly).

height hh=

tt=h/5

Displacement in the first second h=

Displacement h'= in the last second

gt=15t=h=

The conditions you gave are not enough......

Let the height be h, the time to fall to half the height is t, and the velocity is v, then 1 2*h = 1 2*g*t.1)v = g*t

Continue to fall from 1 2h to 1 2h, and the time taken is 1s, then.

1/2*h = v*1 + 1/2 *g*1² =g*t + 1/2*g ..2)

Obtained from (1) and (2).

t² -2t - 1=0

The solution gives t = 1+ 2 (rounding t = 1- 2) so the total time = t + 1 = 2+ 2 (s) by (1) and the total height h = g*t =3+2 2)g

The free fall time of object A is 3s, then the height of A from the ground h1 = gt 2 2 = 10 * 3 2 2 = 45m The free fall time of object B is 1s, and the fall height of A 2 seconds before landing is h2 = 10 * 2 2 = 20m when B.

Free fall s=(1 2)gt 2

The nth second displacement is:

1 2)g(n 2-(n-1) 2) = (1 2)g(2n-1) the displacement in the n-1 second is: (1 2)g(2n-3) so the nth second displacement is more than the previous second:

1/2)g(2n-1)-(1/2)g(2n-3)=g=

It's all nonsense before this question, and if you don't have air resistance, it's standard free-fall.

Then let the falling velocity to the ground be v and the falling time is th=1 2gt 2=v 2 2g v= 2gh and the solution is v=50 2 m s

That's about 71 m s, that's 255 km/h, and you can think of what it would be.

…This obviously does not require university knowledge.

1 Let the continuous equal time be t, 2t, 3t...... respectively

The displacement passed in the first period of time is s1=gt2, s2=g(2t) 2=2gt, s3=g(3t) 2=9gt2

The displacement passed in the first period t1 is s1=gt 2, and the displacement passed in the second period is s2=2gt -gt 2=3gt 2, so the difference between the two periods of displacement is δs=3gt 2-gt 2=gt

You can do the rest yourself, and then count them in turn.

2 I used a v-t diagram to list two equations, let the velocity of the intermediate displacement be v, and the time elapsed from v0-v is t

then the displacements elapsed from v0-v and from v-vt are equal, which is the first equation.

The second equation is that the sum of the two displacements equals the total displacement.

Solving the equation is a bit tricky, but it can be calculated.

3 This rule MS only applies to the period when the initial velocity is zero.

According to s=gt2, there is a ...... s1:s2:s3=t1²:t2²:t3²……

And because of the same displacement, t1:t2:t3 ......=1：√2：√3……This is counted from the beginning).

Thus the first paragraph: the second paragraph: the third paragraph ......=1：√2-1：√3-√2……

4 This is similar to conclusion 2 in that the object passes through the same displacement, t1:t2:t3 ......=1：√2：√3…

The time through s is s=gt 2, so t = (2s g), and the time through 2s can be multiplied by 2 on this basis, that is, the result is obtained.

Finding the time to pass ns is the same as this one.

This question does not belong to the scope of high school students, you first accept this knowledge, when your level of mathematical knowledge and physics knowledge reaches a certain level, you will understand, you will understand, that is, when you go to college, study the knowledge in the book first.