Question

an object is dropped from a building of height 125m. find time taken by object to travel last 1/4 th of its distance ?​

Answer 1

Answer:

Let us first find the total time taken to cover the distance of 125 m.

Using the second equation of motion, we get:

⇒ s = ut + 0.5 at²

⇒ 125 = 0(t) + 0.5 ( 10 ) (t)²

⇒ 125 = 5t²

⇒ t² = 125/5

⇒ t² = 25

⇒ t = √25 = 5 seconds.

Hence the time taken to cover the distance of 125 m is 5 seconds.

Now 1/4th of 125 m is given as:

$\implies \dfrac{1}{4} \times 125 = 31.25 \:\:m$

Distance covered before = Total Distance - 31.25 m

⇒ Distance Covered = 125 m - 31.25 m = 93.75 m

Hence the time taken to cover the last 31.25 m is given as:

⇒ Total Time - Time taken to cover first 93.75 m

Using the second equation of motion we get:

⇒ 93.75 = 0(t) + 0.5 (10) (t)²

⇒ 93.75 = 5t²

⇒ t² = 93.75/5

⇒ t² = 18.75

⇒ t = √18.75 = 4.33 seconds

Hence the time taken to cover the last 31.25 m is given as:

⇒ Total Time - 4.33 seconds

⇒ 5 - 4.33 = 0.67 seconds

Hence the time taken by the object to travel the last 31.25 m is 0.67 seconds

Answer 2

Refer the above attachments for solution.