
- A ball is dropped from a height = H
- The total distance covered in last second of its motion is equal to the distance covered in first 3 second .

- The height of the journey .

We have know that,

Where,
- u = initial velocity = 0m/s
[Note :- Here, acceleration is ‛acceleration due to gravity’ .]
=> S = 1/2 × 10 × (3)^2
=> S = 5 × 9
=> S = <u>4</u><u>5</u><u>m</u> -----(1)
✒ If the ball takes ‛n’ second to fall the ground, then distance covered in nth second is,

=> Sn = 0 + 10/2 (2 × n - 1)
=> Sn = 5 (2n - 1)
=> Sn = 10n - 5 -----(2)
Therefore,
=> 10n = 45 + 5
=> n = 50/10
=> n = <u>5</u>
Now put the value of ‛n = 5’ in equation(2), we get
=> Sn = 10 × 5 - 5
=> Sn = 50 - 5
=> <u>Sn</u><u> </u><u>=</u><u> </u><u>4</u><u>5</u><u>m</u>
The height of the journey is “ <u>4</u><u>5</u><u>m</u> ” .
charged objects will either attract or repel other charged objects
Any force of 29.4 Newtons or greater can do it.
As we know that in transformers we have

here we know that



now from above equation we will have


