If a freely falling body in the last second travels a distance equal to the distance travelled in the first three seconds, the time of its travel is A33 secB44 secC55 secD66 sec

This problem can be solved using the equations of motion.

The distance travelled by a freely falling body under gravity in t seconds is given by the equation:

s = 1/2 * g * t^2

where g is the acceleration due to gravity.

1. The distance travelled in the first three seconds is:

s1 = 1/2 * g * (3)^2 = 4.5g

2. The distance travelled in the last second is:

We know that the total time of travel is (t+1) seconds. So, the distance travelled in t seconds is:

s2 = 1/2 * g * t^2

And the distance travelled in (t+1) seconds is:

s3 = 1/2 * g * (t+1)^2

So, the distance travelled in the last second is s3 - s2 = 1/2 * g * [(t+1)^2 - t^2] = g(t+0.5)

According to the problem, s1 = s3 - s2

So, 4.5g = g(t+0.5)

Solving this equation gives t = 4 seconds.

So, the total time of travel is (t+1) = 5 seconds.

Therefore, the correct answer is C) 5 seconds.