what will be the ratio of the distance moved by a freely falling body from rest in fourth and fifth sec?

Question

please answer this question

Answer 1

The vertical displacement of a body falling freely from rest under the effect of gravity (with no air resistance) is y(t) = 0.5*gt^2, where g is acceleration due to gravity and t is elapsed time. g at the surface of the Earth is 9.8 m/s^2.

Evaluate y for t = 3, t = 4, and t = 5. y(4) - y(3) is the distance traveled during the fourth second. y(5) - y(4) is the distance traveled during the fifth second. Their ratio, i.e., [y(4) - y(3)] / [y(5) - y(4)], is your answer.

Answer 2

using the eq of motion:
s = ut + 1/2gt^2... since its from rest, u=0, thus
s = 1/2gt^2
Now the distance moved during the fourth second will be from t= 3 sec to t = 4 sec
= 1/2g(4)^2 - 1/2g(3)^2
= 1/2g(7)
=7/2 g
Similarly, distance moved during fifth second will be from t = 4 sec to t = 5sec
= 1/2g(5)^2 - 1/2g(4)^2
= 1/2g(9)
= 9/2 g
Thus the ratio would be 7 : 9 or 9: 7

Answer 3

Total distance traveled (y) is given by

y(t)=0.5gt^2,
where g is acceleration of gravity (9.8066m/s/s) and t is time in seconds.

Evaluate this at 3,4,and 5 seconds:
y(3)=4.5g
y(4)=8g
y(5)=12.5g
Distance traveled during fourth second is y(4)-y(3) = 3.5g
Distance traveled during fifth second is y(5)-y(4)=4.5g
The ratio of these two distances is 3.5 to 4.5 or 7 to 9.

Answer 4

Distance = Average velocity x time.

Distance in 4th second = [1/2] [4a + 3a]*1 = 7 [a/2]

Distance in 5th second = [1/2] [5a + 4a]*1 = 9 [a/2].

The ratio = 7/9.