Physics Problems with Free Fall Motion Ask for help 15

1.Let the time taken to start half the journey be t, then (1 2)gt = gt + (1 2)g1 , we get.

t= 2+1, so the total time is t-total = 2+2;height = 2 [(1 2)g( 2+1)].

2.Let the velocities at points b and c be (3 5)v and v respectively, then there is.

v -[3 5)v] = 2gh, h is the distance between BC, so v = 50, so the distance between AC = v (2g) = 125

3.It is easy to find that the velocity of a ball when it falls 1m is v=2 5, and if the fall time of b ball is t, then there is (7-1)+(1 2)gt=2 5t+(1 2)gt, and t=3 5, so the tower height is 16m

4.If the velocity of the ball to the window is v, then the velocity at the bottom of the window is v+, then.

v+, get v=, time is limited, and the rest is calculated by yourself.

Memorize a few formulas, then set the unknowns, "translate" them into equations according to the meaning of the question, and then add the relevant formulas, and you will find that physics is simple.

For example, in the first question, if you ask about the falling time, then set the falling time t. The landing altitude is h, and then read the problem, and the distance that falls in the last second is half of the whole process, and if you find the equality relationship, you can make the equation.

1 2 g t square [total falling distance] - 1 2 g (t-1) square [distance traveled before the last second] = half of the whole journey].

Then write the formula h=1 2gt squared [free fall formula] two equations and two unknowns, you can calculate it (bring the second into the first) The following problems can be like this.

All four questions are drawn first.

1 total height h, then the total time of landing is 2gh under the root number, so the time t experienced before the "last second" is 2gh under the root number and subtract 1, and the displacement that occurs during this time = 1 2 times g and then multiplied by t squared, according to the title, it is equal to h 2, and the rest is to find it yourself.

2 The height between AB is represented by AB, and the height between AC is expressed by AC. The velocity at point B = 2GAB at the root number, the velocity at point C = 2GAC at the root number, and AC=AB+80 find AB first....

3 If we start the study from the time when ball A falls 1m, then the displacement of ball A minus 6 is equal to the displacement of ball B (an equation); a spherical displacement (note that it starts after falling 1m) = the time it takes for this displacement to occur t multiplied by 2g times 1 under the root number, plus 1 2 times g and then multiplied by t squared (two equations); The total displacement of ball b = 1 2 times g and then multiplied by the square of t (again t above), three equations three unknowns.

4 According to the formula "passing through a window with a height of δh=2 m in δt = s" (s = vt + 1 2 times a and then multiplied by t squared) to find the velocity of the ball when it has just reached the top of the window, which is equal to 2 g under the root number multiplied by "the height of the top of the window from the eaves".

...lz, ls are really all in 6, and I won't either.

Classmate, it looks like you're in 6! I'm really a fellow believer, and neither will I

We're all six kinds of people. ==、

Solution: The displacement of the first second h=1 2gt=5m, so the displacement of the last second = 10m, let the total fall time t, then 1 2gt -1 2g(t-1) = 10m, the solution gives t=3 2s, so the total height = 1 2gt=

1 2gt2 = 5 meters. Displacement of 10 meters in the last second.

1 2gt*t-1 2g(t-1)(t-1)=10m, solve t=

1 2gt*t=height, do the math.

bh=t=√2h/g

Substituting data h to find t2:t3 :t4: t5=1:2:3:4 so choose b

As for AB, directly use 1 2gt*t to calculate, CD is simpler, and the calculation of the respective rate is over.

Answer: Displacement of a free-fall body in 0 t seconds: s = h = gt 0 3t seconds, displacement of a free-fall body:

h = g(3t) t 3t seconds, displacement of falling: s = h - h = 4gt 0 6t seconds, displacement of falling free fall: h = g(6t) 3t 6t seconds, displacement of falling:

s₃= h₃- h₂=s₁: s₂: s₃= gt² :

4gt² =1 : 8 : 27

The equation for the motion of a free fall is s=1 2*g*t 2, assuming that the three consecutive journeys are t, 2t, and 3t respectively, then the object is in total motion at the end of the first journey.

1 2*g*t 2, and at the end of the second leg the object has a total motion of 1 2*g*(3t) 2, and at the end of the third leg the object has a total motion of 1 2*g*(6t) 2

The lengths of the three journeys are.

1/2*g*t^2

1/2*g*（3t）^2-1/2*g*t^2=8*1/2*g*t^21/2*g*（6t）^2-1/2*g*（3t）^2=27*1/2*g*t^2

Start free fall s=(gt2) 2;

The first period is 1, s1 = 1

The end time of the second period is 3, s1 + s2 = 9

The end time of the third period is 6, s1 + s2 + s3 = 36 (the numbers of s1,2,3 are simplified, and g and 1 2 are all approximated).

Let the drip time interval be Lingchun t seconds, then the first drop will take 4t seconds to land, the second drop will take 3t seconds, and the third drop will take 2t seconds, which will be obtained by h=1 2gt 2.

h2=1/2g(3t)^2 (1)

h3=1/2g(2t)^2 (2)

h2-h3=1 (3)

solution, t=seconds).

h=1 2g(4t) 2=m).

meters off the ground, when the interval between objects is seconds.

The time taken is 1s, 2s, 3s

The distance passed in the first second h1 = 1 2 * gt 2 = 5 m, the distance passed from the beginning of the second to the end of the third second h2 = 1 2 * g (3 2-1 2) = 40m

The distance traveled from the beginning of the fourth second to the end of the sixth second h3 = 1 2 * g (6 2-3 2) = 135m

1.(1) H=V0*T1+(1 2)*G*(T1) 2 Bring in data t1=1s, h=15m

v0=20m s;

2) to the highest point time t2=v0 g=2s

The highest point is from the throwing point d=(1 2)v0*t2=20m, and the highest point is from the ground h=d+25=45m

The landing time from the highest point (t3) 2=2*h g gives t3=3s

The last total time t2+t3=5s

2 Draw the V-T diagram below.

The green part is the distance between the cars to be kept.

1, (1) Let the initial velocity be V, and the upward direction is positive.

h=vt+(1/2)at^2=15 (a=-10m/s^2)v=20m/s

2) From the throwing point to the highest point, the construction is set to t, and the height is set to hgt=20, t=2s, h=20m

The distance from the ground when the ball reaches the highest point is 20 + 25 = 45

1/2gt^2=45 t=3

So the total time taken is 5s

There is no difficulty in these two questions, and not being able to do them only shows that you are missing some basic concepts

For example, in the first question, you first specify a direction as a positive direction, and then use the formula h=vt+, at this time, when you specify an upward direction, a=-g, you can get v=20m sConversely, a=g, v=-20m sThe plus or minus sign before the number only represents the direction, and since the question asks about speed, you reach v=20m s and the direction is straight up. ......