Urgent!!!!! High one curve movement

According to the title, s1=490m, v1=240m s, v2=25m s, g=because the motion is a flat throwing motion, the partial motion of the motion in the horizontal direction is a uniform linear motion, and the partial motion in the vertical direction is a free fall motion;

s1= 490= t=10s

s2=v1*t=240*10=2400m

s3=v2*t=25*10=250m

Total S = S2-S3 = 2400m-250m = 2150m

It takes time to fall for 9 seconds.

The relative velocity is 240-25=215

That is, the distance is 215*9.

First, calculate the time for the object to move in the air.

1/2gt^2=490

t = root number 10

S = S Aircraft - S Ship = V Aircraft T - V Ship T = 215 * Root No. 10

Set the time for air movement to t

Then the horizontal displacement is vt, and the vertical displacement is 1 2gt 2t=

The velocity along the inclined plane is 8*4 5+12*3 5=

There is a kinetic energy theorem to get s=

Break it down! It is divided into horizontal and vertical or along the inclined plane and perpendicular to the inclined plane and so on.

First of all, it must be that the ball will be able to jump 2mg by force after a quarter of a circumference, and it must be from a quarter point upwards, and the vertical component of the support force will increase from 0 (the later changes are not discussed in a hurry, there may or may not be peaks). It is assumed that the critical condition is reached at a vertical angle a. Using the conservation of energy and Newton's second law, as well as critical conditions, it is obtained.

v0^2=rg(2/cosa+3cosa+2)

In the case of A=Arccos(sqrt(2 3)), v0 has a minimum value v0=sqrt(rg(2+2sqrt(6))).

First of all, I would like to illustrate the idea of this problem: from the conservation of mechanical energy (without considering friction), the square of 1 2MV is equal to the square of one-half MV + mg2r, and the velocity v and vo at the highest point are to be found, and then the force analysis mg=mv2 r is carried out when the ball is at the highest point to find v, and then substitute into the first formula to find vo. When doing such a type of question, it is necessary to conduct force analysis, and not only subjective assumptions.

This is what is shown on the diagram, the m-point, which is an obtuse angle. The force can be decomposed into components perpendicular to the velocity and velocity in the same straight line but in opposite directions, so that the velocity rotates in the direction of the force on the one hand, and decreases on the other. When the velocity turns perpendicular to the direction of the force, the velocity no longer decreases but still rotates in the direction of the force.

In this way, the angle between velocity and force will be less than the right angle and become an acute angle, which is the state of point n. At this time, the component of the force in the direction of velocity is the same as in the direction of velocity. The speed size begins to increase.

Answer the question adds: We know that the direction of velocity always turns in the direction of the force, and as time increases, it will approach the direction of the force infinitely. Even if there is no direction of force, just look at the angle between the initial velocity and the final velocity, if the angle is greater than 90 degrees, you can also know that the force and the initial velocity are obtuse angles.

This is a typical physics question about velocity synthesis and decomposition, the answer is C, the athlete is separated from the target by dv1 v2; The minimum time is d v2

The idea is that the distance to the target is d, but the direction of the speed to hit the target can only be generated by v2 and has nothing to do with v1, so the shortest time is d v2, and naturally the distance between the arrow and the target can be found as v1*d v2, understand?

Thanks for the reference!

The answer is D; First, if we decompose the velocity, we can see that v1 is the straight edge of the right triangle and v2 is the hypotenuse of the right triangle.

So we can find that the velocity of the bow and arrow in the direction of the perpendicular to the runway is (v2 2 v1 2).

Therefore, the shortest time is: d v2 2 v1 2) The shortest distance from the target is: v2 * d v2 2 v1 2) )

If you have any questions, you can follow up!

Choose D, the third year of high school did a bad question.

Solution: The asymptotic line equation for the bivolt rent curve is:

y=(b a)x and y=-(b a)x, (where a=1, b is assumed to be known) concatenate these two equations with y=x+1, respectively, where the values of x,y are x=(-1) (b+1) and y=b (b+1).

Since this point is in the absence of orange trillion y=x+1, the absolute value of the abscissa is equal to the absolute value of the ordinate.

So, the absolute value of b is equal to 1

According to C2=A2+B2

Get c = root number 2

e = c a eccentricity e = root number 2

The result is important, but it is also important to understand the process.

The hyperbolic questions are generally not very difficult, and most of them are calculated, so they are not easy to write.

Hello, the equation can be turned into.

x-1)²/2-y²/4=1

The image of this equation is obtained by translating the image of the equation x 2-y 4 = 1 one unit to the right.

x 2-y 4=1 is a hyperbola, and the value of x can be in the range of (negative infinity, - 2] [2, positive infinity).

Therefore, 2(x-1) 2-y 2=4 represents the range of the abscissa of the curve as (negative infinity, - 2+1] [2+1, positive infinity) and the range of the ordinate y is r

It is obtained by dividing 2(x-1) 2-y 2=4 by 4.

x-1)^2/2-y^2/4=1

Let x-1=h

It can be regarded as h2 2-y 2 4 = 1, that is, hyperbola, so a<- 2 or a> 2

i.e. h=x-1<-2 or > 2

So x<- 2+1 or x> 2+1

And y belongs to r

Let p(m,n) f1(-root(a2+b2),0) c=root(a2+b2).

po^2=m^2+n^2

pf1 2=(m+root(a2+b2)) 2+n 2po 2=pf1 2

m^2=(m+c)^2

m^2=m^2+2mc+c^2

c^2+2mc=0

c=-2m, then m<0

So m<=-a

So e=c a>2a a=2 (m=-a is not true).

The impossibility of the right branch between p and the right branch will not be discussed. Let p(x,y) op= (x y) f(-c,0) pf= ((x c) y ) because op=pf so op =pf so x =x 2cx c gets 2x c=0 -2x=c because p is on the left branch so x -a so -x a 2x 2a because -2x=c c 2a so e 2 eccentricity is [2, ps: all greater than less than signs are heavy rain equal to and less than equal to.