How to quickly calculate the time required from the throwing point to a certain point in a flat toss

According to the meaning of the question, we know that the initial velocity of the horizontal throw is v0, the side length of the small square is l, the gravitational acceleration is g, and the velocity (va, vb, vc) of each point (a, b, c) in the graph is obtained.

Analysis: Let the time from A to B be t, then it can be seen from the graph that the time from B to C is also t (the distance between Ab and B in the horizontal direction is equal).

Since the flat throwing motion can be decomposed into a uniform linear motion in the horizontal direction and a free fall motion in the vertical direction.

So there is 2l v0 * t

Get t 2l v0

The velocity vb at point b is decomposed into two components, horizontal and vertical, and the horizontal component is vbx v0, and the vertical component is vby (l 2l) ( 2t) 3l ( 4l v0 ) 3 v0 4

Vertically: the average velocity over a period of time is equal to the instantaneous velocity at the middle of the period).

So, the velocity at point b is vb root number (vbx 2 vby 2) root number [ v0 2 (3 v0 4) 2 ] 5 v0 4

In the same way, the velocity va at point A is decomposed into two components, horizontal and vertical, so that the horizontal component is vax v0 and the vertical component is vay

Then vby 2 vay 2 2gl

i.e. (3 v0 4) 2 vay 2 2gl

Vay 2 (3 v0 4) 2 2gl

So the velocity at point A is the va root number (vax 2 vay 2).

Root number [ v0 2 ( 3 v0 4) 2 2gl ].

Root number [ ( ( 5 v0 4) 2 2gl ].

Then decompose the velocity vc of point c into two components, horizontal and vertical, and the horizontal component is vcx v0 and the vertical component is vcy

Then vcy 2 vby 2 2g*(2l) (3 v0 4) 2 2g*(2l).

So, the VC root number (VCX 2 VCY 2).

Root number [ v0 2 (3 v0 4) 2 2g*(2l)].

Root number [ 5 v0 4 2 4gl ].

The fastest algorithm should be: knowing the initial speed of the flat throw and the horizontal distance from the point of the flat throw, then you can directly divide the horizontal distance by the initial speed to get the time to this point.

If you don't know the initial velocity and the horizontal distance, but you know the vertical distance, then you can get the t= root number (2h g) according to h=, and you can also get how much time it took to get to this point.

At least 600,000. I don't know if real estate counts.

What else do you know besides muzzle velocity???

Ask!!!

Let the inclination angle of the inclined plane be and the initial velocity of the flat throwing is v0, then when the velocity of the object is parallel to the inclined plane, the distance between the object and the inclined plane is the farthest.

The decomposition velocity, one in the inclined plane direction, the other in the perpendicular direction to the inclined plane, is a uniform deceleration motion in both directions, and the farthest from the inclined plane, that is, the vertical divider velocity is reduced to o, which can be solved.

The object is thrown flat with velocity v0 on an inclined plane with an inclination angle of , and the velocity direction is parallel to the inclined plane at the highest point of motion, and the velocity v=v0 cos and the flight time t=v0cot g

In a flat throwing motion, the initial direction is horizontal and the magnitude is v1. After being thrown, it is only affected by gravity, and under the action of gravity, the object must not only continue to move forward with v1, but also produce free fall motion. If you give the height of the flat throw, you can find the time t to the ground according to h=(1 2)*g*t 2, and it is easy to find how much the object moves horizontally during this time after the time t, and the speed of the landing should pay attention to whether it is horizontal velocity, vertical velocity or combined velocity.

The air resistance is not considered above, if the air resistance is added, the horizontal movement is no longer uniform, and the falling is not a free fall movement, which is a lot of trouble.

When the distance is farthest, the instantaneous velocity is parallel to the inclined plane, according to the slope of the inclined plane and the velocity of the object in the X direction, the velocity in the Y direction can be known at this time, and then the free fall motion from the Y direction can be found.

When the direction of velocity coincides with the direction of the inclined plane, it is the farthest from the inclined plane. You can try to decompose the motion with the bevel direction as the x-axis and the perpendicular bevel as the y-axis, and you can do the work by orthogonal decomposition of the ball velocity and force.

The velocity and distance of the point farthest from the inclined plane in the flat throwing motion, the time - the key point is: the point at which the direction of the velocity of the object is parallel to the inclined plane!

I don't know what to ask!

Can it be proved that the point corresponding vertically is the end of the inclined plane?

What about the figure, there are no other conditions.

Flat throwing motion is a combination of free fall motion and uniform linear motion, and its initial velocity is the horizontal component of velocity. Therefore, the horizontal velocity is calculated as the initial velocity.

This is generally an experimental question, and there will be a few points on the trajectory given. It is sufficient to deduce the horizontal velocity by using the law of uniform variable speed motion. The common inference is that the velocity at the intermediate moment within a certain period of displacement is equal to the average velocity of the distance, and the difference in displacement over a continuous equal period of time is equal to a fixed value:

The product of acceleration and the square of this time.

It can be found according to the height and horizontal displacement; There is a height to get the time, and then the initial velocity is solved according to the horizontal displacement.

Depending on the velocity of a point, it can also be found by decomposing it;

You're not talking about lab questions! Yes, you can use the square t of the vertical x=a to calculate t, and then use the horizontal x=v to start asking if you don't understand.

Altitude h initial velocity v0

Horizontal distance s time t

Vertical direction: h=1 2*g*t 2

Horizontal: s=v0*t

The two equations are solved consecutively.

Let xita be the inclination angle of the inclined plane, assuming that the horizontal velocity of the horizontal throwing is v0 from the top of the inclined plane and the horizontal velocity of the flat throwing is v0 to obtain the component of gravitational acceleration along the direction of the vertical inclined plane a = g*cos(xita), and the horizontal and vertical velocities are decomposed along the direction of the vertical inclined plane to obtain the component of the vertical inclined plane direction. Because of the flat throw, the initial vertical velocity is 0So the initial projection v = v0*sin(xita) +0 = v0*sin(xita).

The motion in the direction of the perpendicular inclined plane can be seen as a uniform deceleration motion with an initial velocity of v and an acceleration of a (negative). v=0 when the leakage is the largest point in place, that is, the farthest point from the inclined plane. 0 - v 2 = 2*a*s , so the maximum distance s = v 2 (-2a) = v0 2*sin(xita) 2 (2*g*cos(xita)).

0 = v +at.

The moment when the maximum distance is reached, t = v (-a) = v0*sin(xita) (g*cos(xita) = v0 g*tg(xita)

The degree of acceleration of the flat throwing motion is unchanged, and according to v=gt, the magnitude of the change in velocity before chaos in the same time is equal and the direction is the same

Therefore, it is selected: Da Biqing A

1. Speed: The Cartesian coordinates of the horizontal and vertical direction of the potato lead can be obtained: vx=v0, vy=gt, and the combined velocity is [v0 2+(gt) 2] under the root number

2. Displacement: x=v0t, y=(gt 2) 23. Acceleration: ax=0 with Cartesian coordinates, ay=g with natural coordinates with an=gcos, a =gsin and sin =gt under the root number [v0 2+(gt) 2]4, trajectory equation:

y=(gx 2) (2v0 2) From the equation, it can be seen that this graph line is a parabola, passing the origin, and the larger the v0, the larger the range.

The direction of the shift of the bits and the horizontal angle of the bundle: tg = sy sx gt 2vo