Option 3 : 8h/9 metres from ground

Electric charges and coulomb's law (Basic)

47904

10 Questions
10 Marks
10 Mins

__CONCEPT__:

- Equation of motion: The mathematical equations used to find the final velocity, displacements, time, etc of a moving object without considering force acting on it are called equations of motion.
- These equations are only valid when the acceleration of the body is constant and they move on a straight line.

There are three equations of motion:

V = u + at

V2 = u2 + 2 a S

\({\text{S}} = {\text{ut}} + \frac{1}{2}{\text{a}}{{\text{t}}^2}\)

Where, V = final velocity, u = initial velocity, s = distance traveled by the body under motion, a = acceleration of body under motion, and t = time taken by the body under motion.

__CALCULATION__:

The ball is dropped from some height (h),

So initial velocity (u) = 0 m/s

Here acceleration is the acceleration due to gravity ( a = g).

Time taken to reach ground = T

Use \({\text{S}} = {\text{ut}} + \frac{1}{2}{\text{a}}{{\text{t}}^2}\)

⇒ \({\text{S}} = {\text{0}} + \frac{1}{2}{\text{g}}{{\text{T}}^2} = h\)

⇒ h = gT^{2}/2

For t = T/3

\({\text{S'}} = {\text{0}} + \frac{1}{2}{\text{g}}{{\text{(T/3)}}^2} = \frac{gT^2}{18} =2h/18 = h/9\)

**Height (H) = h - h/9 = 8h/9**

So option 3 is correct.