Compare the distance a ball falls during the first second?

Question

Compare the distance a ball falls during the first second after it drops with the distance it falls during the second second and the third second.

Answer 1

all object fall 4.9m after 1 sec

X=1/2(9.8m/s^2)(2s)^2
x=19.6m

x=(1/2)(9.8m/s^2)(3s)^2
x=44.1m

Answer 2

The acceleration due to gravity is 9.81 m/s2 in London....Rounding that off to 10 m/s2: it implies, neglecting air resistance, the falling ball bearing will acquire a downward velocity 10 m/s at the end of the first second......20 m/s at the end of second second......30m/s at the end of the third second.

Hence the average velocity during the first second is 0.5 [0 + 10]
= 5 m/s. Therefore the distance fallen in the first second = 5 m

Similarly the average velocity during the second second is 0.5 [10 + 20]
= 15 m/s. Therefore the distance fallen in the second second= 15 m

Likewise the average velocity during the third second is 0.5 [20 + 30]
= 25 m/s. Therefore the distance fallen in the third second= 25 m

It is an arithmetic progression (A.P)....5,15,25,35,45.........with a common difference of 10, which corresponds to the acceleration due to gravity.

Notice that no formula is required here.....understanding is important. The problem can also be solved graphically by plotting a velocity-time graph. Assuming the acceleration due to gravity to be 10 m/s2 helps in the understanding.

Answer 3

We have to make three assumptions to solve this problem simply:
1) we can ignore the effects of the drag of air, and
2) The ball starts out at a downward velocity of zero. That is, the
person doesn't THROW the ball downward.
3) the ball does not hit the ground before the end of the third second!

then the ball's velocity is the rate of acceleration (32 ft/s/s) times the time.
It's easy to picture that in the first second, the AVERAGE speed of the ball will be 16 ft/sec, since it starts out at zero, and finishes the second at 32 ft/sec.
In the second second, the AVERAGE speed is 48 ft/sec.
In the third second, the AVERAGE speed is 80 ft/sec
and so on.
well, 48 is 16 times 3, so the ball falls three times as far in the second second than in the first secone.

For the 2nd-second/3rd-second comparison, the ball falls 80/48 times are far in the third second, or 5/3 times as far.

Answer 4

Since the initial speed is zero, the distnace traveled is given by the formula

S = 0.5 g t^2.

Therefore, S1: S2 : S3 = t1^2 : t2^2 : t3^2,

Substituting t1 = 1, t2 = 2, t3 = 3. We get S1: S2: S3 = 1: 4: 9.

If we want to compare the distances during a particular second only, then we first form the equation for that.

Distance traveled in tn seconds is Sn = 0.5 g tn^2.

Distance traveled in t(n-1) seconds is s(n-1) = 0.5 gt(n-1)^2.

Therefore distance traveled in the nth second alone is
Sn - S(n-1) = 0.5g (2n-1)

Therefore, S1: S2: S3 = 1:3:5.

Answer 5

It would depend on wether you would be taking into account air friction however small. so. say that.... distance = velocity x time.
distance = 9.81 meters per second (gravitational acceleration) x 1 second.... would be 9.81 meters after 1 second.

Answer 6

During each succesive second, the distance relative to the first second's distance is: 1,3,5,7,9,.......(2n-1)

Interesting

Answer 7

hahaha! So authentic. the first time a horse stepped on my foot became very last 12 months, he became a nil.5 clydie, on the right of spring (he took finished good thing about the grass.) And were given truly fat. They needed me to artwork him, so he became huge, 0.5 clydie (huge hooves) 15.3hh, and so overweight! I screamed and screamed and had to throw myself at him to get him to bypass off! large record. like it.

Answer 8

a ball falls at 9.8 meters per second squared...in one second it falls 9.8 meters...in two seconds ..it falls four times as far(39.2m)... in three seconds it falls nine times as far(88.2m)...the first second the ball falls 9.8m, the second, it falls 29.4m....the third second it falls 49 m