It is known that flat throwing is high one physics. High school physics questions in the middle of t

Your condition is not clear, I set the unit of coordinates is meters, the origin of coordinates is at the throw, and the direction of the ordinate is downward (opposite to the usual ordinate).

After the point (20,30) it can be considered that when the object descends 20 meters, the horizontal displacement is 30 meters.

According to h=gtt 2, s=v0t, at (20, 30 points): 20=5tt......（1）

30=v0t……（2）

From (1): tt=4, t=2

Substituting t=2 into (2): the vertical velocity of v0=30 2=15(m s) passing through (20,30) is: v2=gt=10*2=20(m s).

At this time, the combined velocity is: v= 15*15+20*20=25(m s).

The flat throwing motion is the combined motion of the horizontal direction of uniform linear motion and the vertical direction of free fall motion, and this is the idea of solving the problem.

Let the initial velocity v0 and the horizontal direction be s=v0*t

The vertical direction is h=1 2gt 2

where s=20, h=30

The solution yields t=6, v0=20 6

Vertical velocity vh=gt=10 6

So the speed of crossing the point is 100 15

The vertical displacement is 30, and according to h=1 2gt 2, the time of 2h g (20,30) under the root sign is equal to the root number 6, and then according to v=gt. Get 10 times the speed of the root number 6 meters in seconds.

Vertical velocity vh=gt=10 6

So the speed of crossing the point is 100 15

This topic is about the independence of movement! That is, the vertical and horizontal movements are independent! The trajectories of motion are actually their combined motion.

i.e. y=1 2 gt2 x=v0t.。That is, 20=vot, 30=1 2gt2, first find t, then find vo

Wrong. If you throw it flat, the vertical coordinates can only be negative.

Isn't the title a bit wrong? Since it is a flat toss, how is it possible to pass the point (20, 30)?

How can the coordinate origin pass the first quadrant? Is there a mistake?

Note that the inclination angle of the inclined plane is , and the angle between the velocity direction and the horizontal direction of the object falling to the inclined plane is .

According to the displacement relation: tan = 1 2gt vt, calculate t = 2tan v g, and then by the velocity time formula tan = gt v, the synthesis can get tan = 2 tan, you can go.

It's a combination of free fall and horizontal constant motion. It is to look at the ratio of v1 v2 to g to determine the angle.

According to the speed relation:

vx / vy =tan37

vo /gt = 3/4

t =4vo / 3g

According to the displacement relation:

h - sy) /xy =tan37

h - 1 2*gt 2 ) vot = 3 4 substitute t = 4vo 3g into the above equation to get:

h - 1/2* g* (4vo / 3g )^2 =3/4 * vo * 4vo / 3g

vo^2 =9hg/8

vo =3/4 *(2gh)

Solution: Let the angle between v and vt be 1, then it is derived from the title, 1=37°

tan 1=vo vt, vt=gt, h=gt2 2

It can be obtained with VO=3 4 times the root number of 2gh

As can be seen from the figure, the horizontal distance of ab bc is equal, so the time from A to B is equal to the time from B to C, and the vertical distance from A to B is counted, and is set to XL0 The vertical distance from B to C is set to yl0

From the formula δx=a(δt) 2, t=under the root sign ((yl0-xl0) g), then the time is known, the displacement is known, and the initial velocity of the flat throw can be calculated. Don't write it anymore, the one on the first floor seems to be wrong, point A is not necessarily the starting point of flat throwing, the vertical velocity of point A is not zero, and you can't use x=at 2 2 to calculate the time.

Whether the diagram is like this, I don't know if this solution is correct.

After 1 second, vy1=gt=10m s

vy1 vx=tan45 =1 so v0=vx=vy1=10m s

When landing, vt=vx cos60=20m s

vy2/vx=tan60

vy2= 10√3

vy2^2=2gh h=15m

Fall time t=vy2 g= 3 seconds.

x=vxt= 10√3m

take g=10

After 1s, the vertical velocity is 10m s, and the velocity direction is at an angle of 45 degrees to the horizontal direction Horizontal velocity Vertical velocity Initial velocity = 10m s

When landing, the direction of the landing speed is at an angle of 60 degrees to the horizontal direction Landing speed = initial velocity cos60 = 20m s

From v 2=2gh, substituting v=20m s, solving the height h=20m from v=gt, substituting v=20m s, obtaining t=2s; From s=vt, substitute v=10m s, get s=20m

Since the direction of his velocity is at an angle of 45 degrees to the horizontal direction after 1s, the vertical component of the velocity is equal to the horizontal component.

i.e. 1) v0=vy=gt=1*10m s=10m s2) vt==vo cos60=20m s3) vo*sin60=10t, t=sin60 (s), h=(1 2)gh 2=

4) s=vot=10*sin60=5*3^(1/2) m

1。First, determine when the object is furthest from the inclined plane. The distance is defined as the length of the perpendicular line from a point to a line. The velocity at the furthest distance from the inclined plane should be perpendicular to the inclined plane. That is, the direction of the instantaneous velocity at that point is parallel to the inclined plane.

2。Therefore, the ratio of the vertical velocity to the horizontal velocity of the object at this point is tan30 degrees 3 3, i.e., gt vo 3 3, and t ( 3 3)vo g can be obtained

3。Calculate the vertical displacement of the point h (1 2)gt 2 = (1 2) g * (1 3) vo 2 g 2

1/6vo^2/g

4.Calculate the reduction of gravitational potential energy mgh mg(1 6vo 2 g) 1 6mvo 2=1 3e=3j

Let the mass of the object m, the initial velocity be v0, the initial kinetic energy is 1 2mv0 2=9j, mv0 2=2*9=18 When the tangent of the parabola is parallel to the inclined plane, the tangent point is farthest from the inclined plane:

Slope k=tan30°=1 root number 3

When vy vx=k

The horizontal velocity vx=v0 does not change.

Vertical velocity vy=kv0=v0 root number 3

Vertical displacement h=vy 2 (2g)=(v0 root number 3) 2 (2g)=v0 2 (6g).

Gravitational potential energy decreases: mgh=mg*v0 2 (6g)=mv0 2 6=18 6=3j