Chapter 2, Problem 40P

Solution

To show: The distance travelled during each successive second increases in the ratio of odd integers for a free falling object.

Given:

An object is falling freely from rest under gravity

Formula used:

The equations are

  $v=u+at$

  $s=ut+\frac{1}{2}a{t}^2$

Here s is distance , t is time , a is acceleration, u is initial velocity and v is final velocity

Calculation:

As the object is freely falling from rest so for 1^st second the initial velocity is zero.

The distance travelled in first second $s_{1_{}}=0+0.5g$ =0.5g $=\frac{1}{2}g$ ……………….(1)

After falling 1 sec, let its velocity becomes u₁

so $u_1=0+g=g$

So for the next one second its initial velocity will be g , so the distance travelled in the next one second is,

  $s_{2_{}}=g+0.5g$ =1.5g= $=\frac{3}{2}g$ ……………………………………..(2)

Now for the next second , let the velocity is u₂ So $u_2=g+g=2g$

For the next one second its initial velocity will be 2g , so the distance travelled in the next one second is,

  $s_{3_{}}=2g+0.5g$ =2.5g= $=\frac{5}{2}g$ ……………………………….(3)

From equation 1,2 and 3 it is observed that the distance travelled in each successive second, for a freely falling object is increases in the ratio of odd integers