The specific data of the physics flat throwing motion experiment in the first year of high school, t

High School Physics Experiment, Flat Tossing Motion.

Flat throwing object motion experiment:

1. By adjusting the small knobs on both sides of the ball receiving groove, move the ball receiving groove up and down along the strut to place the ball receiving groove on any position on the screen board. (The position should be taken up GT first and moved down one by one).

2. Put the white paper and carbon paper together (the carbon paper is outside) and press them smoothly into the paper clip on the screen board (or use a magnet to press the paper instead). Note that the corner of the rectangular paper should coincide with the origin 0, and its horizontal and vertical edges should also coincide with the two edges of the origin 0.

3. Adjust the position of the ball throwing switch on the track, tighten the locking bolt on the back of the ball throwing switch, load the steel ball into the ball throwing switch, at this time the steel ball will stop at a certain height of the flat throwing guide rail, and then press the ball throwing switch to release the steel ball, the steel ball will roll down along the guide rail groove and be thrown horizontally, fall on the ball receiving groove, and leave marks on the white paper through the carbon paper.

4. Descend the ball receiving groove one by one, repeat the above operation, that is, trace a series of steel ball movement points on the white paper. (Generally take 4 6 points).

5. Take off the white paper and draw the x- and y-axis coordinates of each trace point (the x-axis to the right and the y-axis vertically downward) with 0 point as the origin. The flat trajectory of the steel ball can be drawn by connecting the noted points with a smooth curve.

6. Measure the height h and horizontal distance s of each point, and use the formula h=1 2gt2 and s=vt to find the horizontal fractional velocity of the steel ball, that is, the initial velocity v of the steel ball doing flat throwing motion, and finally calculate the average value.

Glad to answer useful questions for you.

Physical Experiment Report Experimental Topics Studying the Motion of Flat-Thrown Objects Experimental Objectives 1The trajectory of a tracing hail transporting a flat-thrown object. 2.

Find the initial velocity of a flat throwing object. Preparation: Flat throwing motion can be seen as a combined motion of a uniform linear motion in the horizontal direction and a free-fall motion in the vertical direction.

It is only necessary to measure the (x,y of a certain point on the trajectory of the motion from x= to obtain: Instrument Chute, white paper, pushpin, wooden wrench, cardboard card with holes, small ball, hammer line, meter ruler Step 1 Use a pushpin to nail the white paper to the vertical wooden board. 2 Fix the chute in the upper left corner of the board and dig so that the tangent of the end point o is horizontal.

Record the O point on paper, 3 Draw the vertical line of the sail beam through the O point using the heavy vertical line. 4 Press and hold the card with your hand on the flat surface of the board, keep the empty side of the card horizontal, adjust the position of the card, so that the ball rolled down from the groove passes through the hole of the card, and then use a pencil to point a black dot on the notch of the card, which will record the point where the trajectory of the ball is passed. 5 Multiple experiments, tracing multiple points.

6. Connect the points through which the ball passes with a smooth curve to obtain the trajectory of the ball's flat throwing motion. Experimental data 1 Draw the y-axis vertically downward and the x-axis horizontally to the right, using o as the dot. 2. Select four different points a, b, c, and d from the curve, measure their coordinates, and write them down in the table.

According to the formula, the initial velocity of each ball throwing motion is obtained, and then the average value of v0 is obtained. Record x(cm) y(cm) v0(m s) v0 (average).

High School Physics Experiment, Flat Tossing Motion.

[Purpose and Requirements].

Through experiments, it is further clarified that the motion of a flat-thrown object is a combination of the free-falling motion in the vertical direction and the uniform linear motion in the horizontal direction, and the initial velocity of the object is calculated through the trajectory of the flat-thrown object.

Instruments and equipment].

Flat throwing motion experimenter, scale, trace recording paper.

Experimental Method] 1 Installation of flat throwing motion experimenter.

1) Place the flat throwing motion experimenter on the table, install the flat throwing track, and make the projectile section of the track in a horizontal position. Adjust the leveling screws, observe the vertical line, make the panel in the vertical plane, and clamp the positioning plate. (2) The tracing record paper is lined with a copy of carbon paper, which is tightly attached to the recording panel and fixed on the panel with a cardboard, so that the abscissa x-axis is in the horizontal direction, and the ordinate y-axis is downward along the vertical direction, and attention is paid to making the position of the coordinate origin at the place where the steel ball leaves the track.

3) Pull the catch plate to the top one.

2. The steel ball is released close to the positioning plate, and the ball moves downward along the track and is thrown horizontally from the straight part of the track with a certain initial velocity.

3 The falling steel ball hits the catch baffle that is tilted towards the panel, leaving an imprint on the white paper.

4. Pull the receiving baffle down one square again, repeat the above operation method, hit the second imprint point, continue to pull down the receiving baffle until the lowest point, and then obtain a series of trace points when the flat thrown steel ball falls.

5. Remove the recording paper and connect the points recorded in each experiment with smooth curves. The trajectory diagram lines of the flat throwing motion with different initial velocities can be obtained. Their initial velocities can be found by the equations y (1 2) gt2 and s v0t.

Precautions] 1 In order to ensure the accuracy of the experiment, it is necessary to ensure that the recording panel is in the vertical plane, so that the plane of the flat throwing track is close to the plate surface.

2. When placing the blank paper, the origin of the coordinates should coincide with the throwing point of the projectile, so that the starting point of the movement trajectory of the projectile can be correctly determined, so as to determine the x and y coordinates of any point on the trajectory.

High School Physics Experiment, Flat Tossing Motion.

s=v0t+1/2gt^2

In the horizontal analysis, it takes t to fall from C to D, and 2t to fall from C to E, and the vertical direction is for the column formula from C to D and from C to E, v0 is the horizontal velocity of point C 30-15 = v0t + 5t 2

40-15=v0*2t+5(2t)^2

You can solve v0 and t.

Solving this problem is the key to equal the difference. The vertical displacement from A to C is, from C to E, according to the difference formula (the difference of the displacement in adjacent equal time is constant) x AT2, yes, the solution T , then Vo can be found from C to E, and the velocity of point C in the vertical direction is V, then the time taken from O to C is t 2 10, so the coordinates of point O are (-20, -5).

Let each small time interval be t, the falling time of point B t, and the vertical distance between point A and the falling point is x5(t2)=

5((t+t)2)=

5((t+2t)2)=

The solution yields t=then v=1

t=x=coordinates:(, the 2 after parentheses are squared.

Let the coordinates of point o be [x,y] and t seconds elapse when point o moves to point b.

x+10=1/2gt^2 x+15=1/2g[t+t]^2

y+10=vt can be solved to get t v x y

Let the horizontal initial velocity be U and the vertical velocity at point A be V, then there is:

AB time: t1=x1 u

AC segment time: t2=(x1+x2) u

Vertical displacement of AB section: V*T1+1 2G*T1 = Vertical displacement of Y1AC section: V*T2+1 2G*T2 =Y1+Y2 Solution:

u=√((g*x1*x2(x1+x2))/(2(x1*y2-x2*y1)))

There is a formula for the flat throwing motion, knowing x1, x2, y1, y2, you can list the formulas and then synthesize each other, so that you can calculate the velocity of the place.

High school physics studies the flat throwing motion of an object.

How do you let others help you?

The flash frequency is 30Hz, so the interval between bright spots is t=1 30 seconds Longitudinal displacement h=h=gt 2 The solution gives g=

Transverse displacement s=3 Time t=1 30 seconds initial velocity v=s t= At point a, time is 2t, so the longitudinal velocity is 2gt= transverse velocity = initial velocity.

Thus point a and velocity are under the root number: =

Hello classmates, I am Cui Yan, a teacher from New Oriental Youneng Learning Center.

This problem can be thought of in this way, the flat throwing motion is characterized by a uniform linear motion in the horizontal direction, and a free fall motion in the vertical direction.

The vertical distance from point C to point B is the side length of a small square, and the side length of three small squares from B to A is the vertical distance ratio.

1:3, so c is the initial throw point, and the time from c to b is 1 30s, by.

H = gt 2 2 , g = 2 h t 2 =

The initial velocity v=x t=3*

The velocity at point a is: the square of the horizontal velocity + the square of the vertical velocity and then open the root number.

The vertical velocity is vy=g*t (t is the time from the movement of c to a) to obtain:

Finally, the velocity of point A can be found as:

Good luck.

Solution: The flash frequency is 30Hz, so the interval between bright spots is t=1 30 seconds.

So: t=1 30s l= h=hcb -hba x=3l by the uniform acceleration kinematic formula:

h=gt² ①

h=hcb -hba ②

Lianli understands:

g = by the kinematic formula:

v0=△x/t

Substituting the data to solve:

v0 = point a velocity: vy = gt

v0=△x/t ②

Simultaneous solution: va = root number vy + v0

The horizontal displacement from C to B to A is equal, i.e., the time taken is equal.

In the vertical direction, use h2-h1=gt2 to do this. That is, the difference in displacement in a continuous equal time is equal to a constant quantity (at2).

g=h2-h1 t2, t=1 30s h2=, h1= , get g=second space, v=x t x=bring in v=

In the third space, the velocity of a is required, and the vertical velocity of b vby=(h1+h2) 2t= is first obtained

Then the law of motion of uniform velocity in the vertical direction: vay = vby + gt = the velocity of a = the square of vay under the root number and the square of the initial velocity on the root number are obtained.

t=1/30

Because h=(1 2)gt 2

The solution gives g = horizontal velocity v = 3l t =

a, with vector synthesis, vertical velocity v'=g*2t=

Root number (v 2 + v.)'^2)=

The square side length is.

By the title, t = 1 30s

The horizontal distance between C and B, B and A is equal, and the flat throwing motion is uniform in the horizontal direction, so the time of C and Ba is also equal.

by x=gt 2, x=h(ab)-h(cb)=3*g= x t 2=

Muzzle velocity v0=

Vertical v(a)=gt=

v(a) = under the root number.

Transverse displacement s=3 Time t=1 30 seconds Initial velocity v=s t=

14.Figure 12abc

x/cmy/cm

o In the experiment of studying the flat throwing motion of the ball, a student recorded three points A, B, and C (A is not the throwing point), and established a coordinate system as shown in Figure 12, and the coordinate values of the three points on the flat throwing trajectory: the coordinates of point A are (5, 5), the coordinates of point B are (15, 20), and the coordinates of point C are (25, 45). Take g=10m s2, then the initial velocity of the ball is m s; The coordinates of the point where the ball is thrown are x= cm, y= cm.

Figure 12abcx cm

y/cmo

It's not a difficult question. The horizontal direction of the flat throwing motion is a uniform motion! aa is a constant velocity, and bb is also a constant velocity. You have to make sure that the time from the throw point to a is t1, and the time from the throw point to b is t2. The height difference of the fall is as shown in the title h=...

Bring in two times to solve the initial velocity.

It is to use the vertical direction of s=gt square to calculate the time, and then you can find the transverse velocity, and then you can find all kinds of things.

Isochronism calculations using horizontal and vertical motion.

Record the trajectory on a piece of paper with small squares printed on it, and the side length of each small square is l = cmIf the positions of the ball in the flat toss motion are points a, b, c, d in the figure, then the initial velocity of the ball is calculated as v0= (denoted by l and g), its value is , and the velocity of the ball at point b is(take g = m s2).

Analysis] According to the knowledge of kinematics, it can be known that v0=2l t

s=gt2 ②

where s=3l-2l=l, so , substituting is obtained.

The vertical partial velocity of point b (using the law that the instantaneous velocity at the intermediate moment is equal to the average velocity of the whole time).

The rate of point b.

High School Physics Experiment, Flat Tossing Motion.

For a uniform acceleration linear motion with an initial velocity of 0 in the vertical direction, s=gt2 2Then the ratio of the distance traveled by the object in the same time interval is 1:3:5:7...

Observe ab:bc (vertical height) = 3::5, from which it can be seen that the time between ab and bc is 2t and 3t, where t is the time interval of taking pictures.

The vertical height of ab is 15cm, that is, ((2gt) 2-(gt)*2) 2g=3g t 2 2=15cm, and the time interval t= can be found

The horizontal direction is a constant movement, moving 3 blocks at a time. From the obtained time interval, the horizontal velocity v level = 15 cm t = can be obtained

The vertical velocity is 2gt=2m s

The velocity is the sum of the squares of the horizontal and vertical velocities v=