High School Free Fall Problem!! High School Physics Free Fall Problems, Ask for !!!!!!!

It should be the wrong answer.

It takes seconds for a ball to fall from a height of 100 meters (that is, 20 under the root number), and if you divide it, you can throw 8 more balls in 4 seconds, so there are 9 balls in the air.

From h=1 2gt2(h=100m, g=10m s2), t=《So the number of balls n=(

To g=10 The ball takes about 20 seconds to fall from the top of the 100-meter tower, and it is not higher than the second.

Take 4 seconds and divide it by 8, and you either add an extra one or g=

The time should be, so it should be 9+1=10.

Your algorithm is correct, but this problem has already released a ball in the first instant, so you have to add one! The specific process is as follows:

1/2gt^2=100m

Got: T root number 20

The number of time intervals is: t

Since the balls are released at the beginning of time, the number of balls in the air is the time interval plus 1, (like from one to three, with only 2 intervals in between) then the answer is 9

That's 8. The 9th one is still in your hand, and the 9th one will be released after the first moment it lands in seconds.

The answer is right. The time t it takes for a ball to land from 100m can be found by h=1 2gt 2.

t=root, 20=

When you throw the ninth ball, the first ball you throw has already landed. So it can only be eight, not nine. Rounding is not possible here.

The volume of water flowing out of the faucet per unit time is equal.

Due to the free fall, the water velocity is greater the lower it is (i.e., the longer it moves), so there is a displacement difference with the water at the next moment.

But as long as you grasp the equal volume of water flowing out per unit time, you can equivalently see the new water flowing out at the bottom, and the position and shape of the previous water are unchanged, so it is more intuitive and easier to understand.

v1=1m/s

v2=1+(2gh)^(1/2)

s1:s2=d1^2:d2^2

v1*s1=v2*s2

Note: x y is the meaning of x to the y power.

1 。The rope is straightened, i.e., the displacement difference h=ha-hb= m i.e. - tb) 2 = m

ta =tb+

Substituting i.e. solution TB ,ta

Velocity va = g ta

ha = （ta）^2

2.Same as 1.

g（ta）^2 - = 25 m

ta=tb+△t = 2 s + t

Solve the equation to get t =1 s

1.。Let the B ball be released for t seconds, and the rope will be straightened, and their spacing will be the length of the rope, 1 2g (t + square - 1 2gt square =

There is 1 2g (t + square - 1 2at square =

Solve the t below and you'll be there.

2..Similar to the question above.

1 2g (t + 2) squared -1 2g2 2 = 25 you understand.

When the object is ascending, the velocity becomes zero in three seconds, so at the end of five seconds the object is in the process of falling, so at the end of the sixth second, it returns to its original position.

=2。The displacement is equal to the displacement of the 5th 6 seconds after departure = 25m3. v = velocity at departure - velocity at the end of the fifth second = 30-20 = 10m s4. Average velocity = displacement time = 25 5 = 5m s

In the vertical upward direction, the object goes up in 3 seconds, and the velocity becomes zero, so at the end of 5 seconds, the object is in the process of falling, so at the end of the 6th second, it returns to its original position.

==25m3。v = v end - v beginning = (—20) - (30) = —50m s4. Average velocity = displacement time = 25 5 = 5m s

Solution: Method 1: Object: Whole process, h=

Object: h=h-200 process, h=

The simultaneous solution is t=7 s and h=245 m

Method 2: H1=200m process, the instantaneous velocity at the midpoint of time is V1=H1 T1=50 m s

Object: The process from the beginning to the midpoint of time, v1=gt, gives t=v1 g=5 s, so the whole time t=t+ s

The total height h= m

The initial velocity at the last 200m can be calculated according to s=v zero t + gt square 2.

Then according to v=vzero+gt, the velocity at the time of landing can be calculated.

Then v=gt calculates the time; v square = 2gh to calculate the height.

h=gt^2/2

9/25h=h-1/2g(t-1)^2

Subtract the two formulas to obtain 16 25h=1 2g(t-1) 2=16 25 1 2gt 2

Find t and then h

Solution: h=1 2gt 2 1

9 25h=1 2gt 2-1 2g(t-1) 2 2 simultaneous solution:

t = 5 or 5 9 (discarded because it is less than 1).

h=125m

That is, the tower is 125m high

If you draw the ST diagram, you can see that it must have met when the first ball fell and the second ball went up. Then it takes 3s for the first ball to be thrown to 15m (the formula for the destruction of the town is solved) s=vot+1 2at2, and the second ball takes (5-13) 2 to rise to 15m

3. Subtract the second result to get the answer sock filing.