A physics problem about finding the coordinates of a thrown point

This one is simple: set the position of the throw point to o. Let the time taken from o to c be t, and the time taken from o to b is based on known conditions.

Then: the distance from o to c in the vertical direction y=

In the same way: the distance from o to b in the vertical direction z=Note: Both equations are based on the formula: y= and then the vertical distance between z-y= is solved to obtain t=

Then bring t into the equation y=get that the vertical distance of o from c is o to c, so o to a is used.

You can calculate the horizontal distance from a as.

Total: o m above a, m on the left.

Because the horizontal coordinate difference between AB and BC is equal, and the horizontal motion is a uniform motion, the time A to B and B to C are equal. And because the vertical direction is a uniform acceleration motion with acceleration g, the vertical displacement difference between AB and BC is equal to GT 2, i.e.

The solution is t=, and b is the midpoint time of ac, so its vertical velocity is the average velocity of ac in the vertical direction, that is, the time from the throwing point to the b point is calculated from this velocity, and the vertical displacement is, so the ordinate of the throwing point is 15cm-20cm=-5cm

The horizontal velocity (i.e., the average velocity in the horizontal direction of AC) v=, the horizontal distance from the throwing point to a is 1m s*, so the abscissa of the throwing point is -10cm.

The initial velocity and time interval you gave in this question can actually be calculated, so you don't need to tell it.

Remember the standard unit, ah......It should be centimeters.

Point B is separated longitudinally from point C.

According to the displacement formula: x=v0t+ where x=,v0 is the partial velocity in the longitudinal direction of point b, a=g=10m s2,t=

Solve the equation and get vb=2m s 2

The longitudinal velocity v=g*t, the throwing time is before point b.

In this second: lateral displacement x=; Longitudinal displacement y=

So throw the point coordinates (-10,5).

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For: <>

Passing the fixed-point <>

Remember the standard unit, ah......It should be centimeters.

Point B is separated longitudinally from point C.

According to the displacement formula: x=v0t+

where x=,v0 is the partial velocity in the longitudinal direction of point b, a=g=10m s 2, t = solve the equation, and vb=2m s 2

The longitudinal velocity v=g*t, the throwing time is before point b.

In this second: lateral displacement x=; Longitudinal displacement y=

So throw the point coordinates (-10,5).

Solution: From the graph, it can be seen that the vertex coordinates of the parabola are (2,-1), so the analytical formula of this parabola can be set by the vertex formula as:

y=a(x-2) 2--1, and it can be seen from the figure that one of the intersections of this parabola and the x-axis is (1,0), so o=a(1--2) 2--1

0=a--1

a=1 So the analytic formula for the parabola is: y=(x-2) 2-1, and the general formula is: y=x 2--4x+3

Let the parabolic analytic formula: y=ax 2+bx+c

According to the title:

c=2a+b+c=0

9a+3b+c=0

The solution is: a=-2 3, b=-4 3, c=2, y=-2 3x 2-4 3x+2

That question, that question.

4) Flat throwing movement: There is a falling body movement in the vertical direction.

ybc-yab=gt^2=

t= f=1/t=10

v0=3l/t=

vby=(8l/2t)=2m/s

vb=(v0^2+vbt^2)^1/2=

vby=gt t=

Time to reach point A t1=

x=v0t= y=1/2gt^2=

Coordinates (

Solution: Let the analytic formula of the quadratic function be y=a(x+3) 2 and substitute the points (1,6) to get a= <>

The analytic formula for parabola is y= <>

x+3） 2

Substituting the points (1,6) into the line y=2x+m gives the function relation of the line m=4 as y=2x+4

Solution: y=- <

x+2) Dig Zhenghui judgment answer 2

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Put away the <>

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