Question

When dropped from rest, any object will fall a total distance of 19.6 m in the first 2 seconds of its fall. How fardoes the object fall just during the 2nd second (i.e., from time t= 1 s to t = 2 s)?

Answer 1

secondThe distance covered by the object while falling, s=19.6 m

Time taken by the object to cover the distance, t=2 s

As the object is dropped from the rest, its initial velocity is u=0 m/s

The acceleration of the body will be given by the formula,

`s=ut+(1)/(2)at^2`

Where a is the acceleration of the object.

On rearranging the above equation,

`a=(2(s-ut))/(t^2)`

On substituting the known values,

`a=(2*(19.6-0))/(2^2)=9.8\text{ m/s\textasciicircum{}2}`

The velocity of the object after t=1 s can be calculated using the formula,

`v=u+at`

On substituting the known values,

`v=0+9.8*1=9.8\text{ m/s}`

This will be the velocity of the object after one second. This will be the initial velocity of the object at the beginning of the 2nd second.

Thus, the distance covered by the object during the 2nd second can be calculated using the equation,