a free falling object falls a greater distance during each second than the total distance fallen in all the?

Question

previous seconds. True or false? and why

Answer 1

Well, if we factor in terminal velocity, it is false because then it will stop accelerating so it travels the same distance each second.

However, if we have an object having fallen for 5 seconds, it will be going about 50m/s, and the next second it will be going about 60m/s. Previously, it will have travelled the integral of 10x^2 (10 * .3333x^3) from 0 to 5, it will have traveled 408 meters, but in integral from 5 to 6, it will only have travelled 297.2 meters. So that proves it is false.

Answer 2

Nope.
The distance an object falls in one second is .5*32*1^2 = 16'
2nd second = .5*32*2^2 = 64'
3rd second = .5*32*3^2 = 144'
4th second = .5*32*4^2 = 256'
5th second = 400'

The 2nd second is more than all previous seconds combined.
64 > 16
The 3rd second is more than all previous seconds combined
144 > 64+16=80
The 4th second is more than all previous seconds combined
256 > 80+ 144 = 224
But the 5th second is less than all previous seconds combined
400 , 224+256 = 480

This is true even without atmosphere to consider!

Basically, the question is, is x^(n+1) > Sum(x^n + x^(n-1) + ... x^1 + x^0) ? The answer is no.

Answer 3

The acceleration by way of gravity is 9.80 one m/s2 in London....Rounding that off to 10 m/s2: it implies, neglecting air resistance, the falling ball bearing gets carry of a downward velocity 10 m/s on the conclusion of the 1st 2d......20 m/s on the conclusion of 2d 2d......30m/s on the conclusion of the third 2d. for this reason the traditional velocity in the process the 1st 2d is 0.5 [0 + 10] = 5 m/s. subsequently the gap fallen interior the 1st 2d = 5 m in addition the traditional velocity in the process the 2d 2d is 0.5 [10 + 20] = 15 m/s. subsequently the gap fallen interior the 2d 2d= 15 m Likewise the traditional velocity in the process the third 2d is 0.5 [20 + 30] = 25 m/s. subsequently the gap fallen interior the third 2d= 25 m that's an arithmetic progression (A.P)....5,15,25,35,forty 5.........with an undemanding distinction of 10, which corresponds to the acceleration by way of gravity. word that no formula is needed right here.....awareness is substantial. the subject is often solved graphically by way of plotting a velocity-time graph. Assuming the acceleration by way of gravity to be 10 m/s2 facilitates interior the awareness.

Answer 4

for a free falling body the forces on the body are its weight & friction so F=mg
since mass constant & g also
therefor f cosnt. & distance is not greater ............
false

Answer 5

False, off the top of my head. Especially when you consider terminal velocity.

Answer 6

true, due to gravity the object is constantly accelerating meaning is velocity is incresing amd if the object is moving faster it is movin a greater distance in a period of time