Whoever gives us a set of physics questions for the first year of high school, the physics questions

High School Physics Free Fall.

What is the time it takes for a free-fall body to fall half of its height if the physical landing speed is vt? How long does it take to descend half the height?

It's the first half of the time. and the second half of the time).

Answer: According to the displacement-time image, the red part is half of the displacement, and the similarity ratio can be obtained.

t1 2: t total = 1: root number 2, v1: vt = 1: root number 2t total = vt g = vt 10

t1 2 = vt (10 root number 2).

After v = (root number 2-1) vt (10 root number 2).

The time required to drop half the height is vt (10 roots 2), the time required to fall half the height (roots 2-1), and vt (10 roots 2) <>

If the velocity of an object thrown in the horizontal direction is 10m s, how long does it take to find the direction of the combined velocity at an angle of 60 degrees horizontally, and what is the combined velocity at this time? What is the displacement of the motion of an object?

A, B and C move through the same displacement, and A accelerates at a1; B first accelerates uniformly with a2>a1, and then a3a1. Q: Which of the three will pass through this displacement first (i.e., by the size of the equal displacement t A, t B, t C)?

The answer is always tB, tA, tC.

See if there's anything you want.

The first question is the old college entrance examination question, according to the time equality of the AB segment and the BC segment, we associate it with the dot timer paper tape information processing method.

That is, vb=(l1+l2) 2t l2-l1=at 2, and then sob=vb 2 2a soa=sob-l1 and calculate it yourself.

Question 2. The time difference is 6 seconds, which is the time it takes from the lower layer to the upper layer and back again.

That is, from the lower layer to the upper layer it only takes 3s

Direct s=3*1 3=1km

Let the buoyancy force be f (because the volume is unchanged, let us consider f to be a fixed value). According to the knowledge of acceleration, it can be easily found that the acceleration of uniform acceleration is 1 meter per square second, and the acceleration of uniform deceleration is meter per square second. Let the weight of the lost object be x.

According to Newton's second law, (m1g-f) m1=a1, (f-m2g) m2=a2 (a1=1,a2=,m1=990,m2=990-x,g=, solvable, x=101kg

2.I think my approach to this problem is quite stupid, but I don't know how to do simple algorithms.

The way to verify that the ball will fall on that step is to find t using h=1 2gt and then find the distance from x=vt.

Take the first section of the step as an example, find t=, and then find x=, so it can't fall on the first section of the step, and then count in turn until x y (y is the width, the first section is, the second section is, it can be obtained, and it falls on the third section of the step.

Question 1: 101kg

mg+mg-f=990*1

f-mg = (990-m) acceleration of deceleration motion) is solved to obtain m=101kg

Question 2: If the platform you start on is level 1, you will be on level 4.

The condition for smashing the ground is sy>sx

When you get t = is the conversion point, it is impossible to smash it before, and then it must be smashed on the first stone level passed.

Then calculate the sx of the time, and the ball is above the four.

1.Let the downward acceleration of the balloon after the air leakage is a, so there is 1m s*4s+1 2*a*(4s) 2=12m

The solution is a=1m s

At this time, the upward traction force is f=mg-ma (m is the mass of all things) and the mass of the pressure chamber is m', the upward acceleration is a'

Hence the a'=(3m/s)/5*60s f-(m-m')g=(m-m')a'

Bring in the data to calculate and get m'。

2.（1/2)gt^2=

s=Bring in the data to calculate.

1: When the balloon is blocked, the descending speed is 1m s, and the uniform acceleration movement is done, and it drops 12m within 4s, so the downward net force is calculated: f=ma; s=vt+1/2at*t; f=990kg*1m/s-2=990n；The buoyancy of the repaired balloon is obtained:

ffloat=mg-f=;

The velocity decreases by 3m s in five minutes, the acceleration is a1=, the direction of the resultant force is upward, and the throwing mass is m1, then: f float - (m-m1)g = (m-m1)a, from which m1 = 101kg is obtained

2: After the ball is kicked out, it is a flat throwing motion: it satisfies s=vt in the horizontal direction, and it is a free fall motion in the vertical direction, satisfying:

s=1 2gt*t, the ladder is equal in width and height, then: it can be obtained by the two formulas, t=2v g=take g=10m s-2), that is, it falls first on the third order downward.

2gs=v 2, s=20m, v = 2gs open root number = 12m s

20+25=45m

v to find the slag resistance file = 2gs, the root number = 30m s

When the first half of the train is accelerated and the first half is slowed down, there is a minimum of chaos.

2as=v^2,v=at

v=28m/s,t=20*2=40s

72km/h=20m/s

Do not make the two cars collide, just equal speed.

v2'=v2+at

That is, 8=20+at.........a

x1=x2 x2=v2t+at^2=20t+

8t+600m=20t+

Combined with AB, t=100s,,a=

Solution: From the meaning of the question, the two cars are moving in the same direction, if the truck is used as a reference, then the speed of the passenger car is v=, and the solution from v 2=2as can be solved a>.

If you choose BA wrong, the kinetic energy will increase first and then decrease.

B Yes, note that the mechanical energy of the ball is reduced, and the mechanical energy of the system composed of the ball + spring is conserved, and it is not conserved.

Don't be fooled by words.

c, d is false, the decrease in the kinetic energy of the ball + the gravitational potential energy is equal to the increase in the elastic potential energy of the spring.

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The velocity ratio is 1:2 and the kinetic energy ratio is 1:4).

6.The elongation is 1n 500

f=,v=18km/h=5m/s

p = average speed of 100

a=2s/t^2=