How many seconds does it take for a ball to fall to the ground from the height of 10 meter? (g = 9.8 1m/s 2, ignoring air resistance)

The process of error method calculation:

Among them,

Because:

So:

Finally: keep two decimal places, t= 1.43.

The core of calculating this problem is:

Let me be clear,

The data in the last bracket is obviously between 1-0.0005 and 1+0.0002, while the root number two is between 1.4 135 and 1.4 145, so it is+. 1.01* (1+0.0002) *1.4145, so 1.426

Extended data:

How to open a square root

Take the number 748726784 1 as an example. First, divide this number into digits, and divide every two digits into 74, 87, 26, 78, 4 1. This * * has five paragraphs, which means the result is five digits. From left to right, the calculation method of the first digit is to find the maximum number (8) whose square does not exceed the first segment number (74), then? And then the square of the difference from the original number,

Then the next calculation method of each bit is to multiply the result by 2, then find the largest number, let this number add the product just now and multiply it by the difference that is not more than before, and then find the difference. , ? (80000*2+6000)*6000=996000000; 108726784 1-996000000=9 126784 1;

, ? (86000*2+500)*500=86950000; 9 126784 1-86950000=50 1784 1; ; (86500*2+20)*20=3460400; 50 1784 1-3460400= 155744 1; ; (86520*2+9)*9= 155744 1; 155744 1- 155744 1=0; So you may be a little surprised that we got the result. Why is this? Why can you calculate the root number like this? In fact, the fact is very simple, that is, Newton's binomial theorem: (a+b) 2 = a 2+2ab 2 = a 2+(a * 2+b) * b, where a is the result obtained in the calculation: 86000, b is the next number to find:? ? Five of 86500.

This method is cubic root, fourth power,? When n=3, (a+b) 3 = a 3+3a 2b+3ab 2+b 3, then if we ask for a cube root, we only need to find the maximum number b once (the result is a), so the value of 3a 2b+3a 2b 2+b 3 is smaller than that obtained before.

So ... I use multiples of integers as an example. You might as well try any number, and then calculate it according to the condition of how many significant digits are reserved. Keeping a few significant figures means calculating several times. This method is always effective.