A stone falls freely such that the distance covered by It in the last second of its motion is equal to the distance covered by it in the first 5 s. It is in air for:

Question

A stone falls freely such that the distance covered by It in the last second of its motion is equal to the distance covered by it in the first 5 s. It is in air for: (A) 26 s (B) 25 s (C) 13 s (D) 12 s

Tardigrade Admin · Accepted Answer

In case of freely falling body initial velocity is zero.
From equation of motion the distance travelled in the

n

th second of motion is

s_{n} = u + \frac{1}{2} g (2 n - 1)

....(i)
Also

s = u t + \frac{1}{2} g t^{2}

.....(ii)
Putting

u = 0

in Eqs. (i) and (ii)
and

t = 5 s

in Eq. (ii), we get

s = \frac{1}{2} g (5)^{2} = \frac{25}{2} g

,

s_{n} = \frac{1}{2} g (2 n - 1)

Given,

s_{n} = s

∴ \frac{1}{2} g (2 n - 1) = \frac{25}{2} \times g

\Rightarrow 2 n - 1 = 25

\Rightarrow n = 13 s

Q. A stone falls freely such that the distance covered by It in the last second of its motion is equal to the distance covered by it in the first 5s. It is in air for:

26 s

25 s

13 s

12 s

Q. A stone falls freely such that the distance covered by It in the last second of its motion is equal to the distance covered by it in the first $5 s$ . It is in air for: