A freely falling body travells ------- % of total distance in its 5th second is ?

Question

Ans: a)8% b)12% C)25% d)36%

Answer 1

(5^2 - 4^2)/5^2 = 9/25 = 36/100 = 36%

Edit:

The distance travelled by a falling body is described by the equation

s = s0 + v0t + (1/2)at^2

In this case s0 = 0 and v0 = 0, so

s = (1/2)at^2

The distance the body falls in 4 seconds is

s(4) = (1/2)a(4)^2

The distance the body falls in 5 seconds is

s(5) = (1/2)a(5)^2

so the distance the body falls in the 5th second is

s(5) - s(4) = (1/2)a(5)^2 - (1/2)a(4)^2

Factoring out the (1/2)a,

s(5) - s(4) = (1/2)a[(5)^2 - (4)^2]

Dividing by the total distance,

[s(5) - s(4)/s(5) = (1/2)a[(5)^2 - (4)^2]/[(1/2)a(5)^2]

The (1/2)a terms cancel out, leaving

(5^2 - 4^2)/5^2 = (25 - 16)/25 = 9/25 = 36/100 = 36%