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A simple physics problem (particle motion) At the same position above the ground, two small balls are thrown upward at the initial velocity of V1V2. After the second ball is thrown, it meets the first ball in t time. By changing the time interval between the two balls, the T value can be changed. Let V1V2 be selected and V1 be selected
If two objects fall, they can be regarded as having zero initial velocity and the same acceleration as G, but it can be regarded as the pursuit with time difference, which is obviously impossible to catch up (this also excludes meeting below the throw point). The rest is 1. Both balls rise 2. One rise and one fall
There is a method called "fast shot" in the movie stunt. For a car that starts to do uniform acceleration straight-line motion from static state, when the stunt is not used, the acceleration of the car on the screen is a, and the speed when the car moves to a certain point is v. when the double speed "fast shot" is used, the acceleration of the car on the screen and the speed when the car moves to the same point are ()
A. 2a、2vB. 2a、4vC. 4a、2vD. 4a、4v
Because the displacement of the object on the screen is the same when using the "fast lens" as when not using the "fast lens", according to the formula a = v22x, if x is the same, then A2 = 4A1
A tennis player hits the ball with a racket to make it fly out horizontally. After the first ball falls on his own court, it bounces up, just rubs the net and falls on point a of the other court. As shown in the figure, the second ball rubs the net directly and also falls on point A. suppose that the collision between the ball and the ground is a completely elastic collision, and the air resistance is not taken into account What is the ratio of height h to net height h?
According to the law of flat throw, the time from the first shot to the first landing is equal. Because the collision between the ball and the ground is completely elastic, suppose the time from the first shot to the landing is T1, and the time from the second shot to the landing is T2, then: T1 = 3t2
When an object is thrown horizontally at an initial velocity of 5 m / s, its horizontal displacement and vertical displacement are equal when it lands. (g takes 10 m / S2)? (2) What is the actual displacement of the object in this process? (3) What's the landing speed?
(1) According to 12gt2 = v0t, t = 2V & nbsp; 0g = 2 × 510s = 1s (2) horizontal displacement x = v0t = 5 × 1m = 5m, then the actual displacement of the object is s = 2x = 52m. (3) vertical velocity of the object: vy = GT = 10m / SV = V20 + v2y = 55m / s. A: (1) the object lands after 1s. (2) the actual displacement of the object in this process is 52m. (3) the velocity of the object when it lands is 55m / s
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