Question

Two bodies begin a free fall from the same height at a time interval of N s. If vertical separation between the two bodies is 1 after n second from the start of the first body, then n is equal to
(a) $\sqrt{nN}$
(b) $\frac{1}{gN}$
(c) $\frac{1}{gN} + \frac{N}{2}$
(d) $\frac{1}{gN} - \frac{N}{4}$

Answer 1

Answer:

C) 1/gN+N/2

Explanation:

Height of two bodies = h (Given)

Time interval - n seconds (Given)

Vertical separation between two bodies = After 1n seconds (Given)

s1 = 1/2gn²

s2 = 1/2g(n−N)²

s1−s2=12g[n²−(n−N)²]

1=g/2[2n−N]N

n = 1/gN+N/2

Thus,  If vertical separation between the two bodies is 1 after n second from the start of the first body, then n is equal to 1/gN+N/2

Answer 2

Answer:

Let x

1

be the distance travelled by the first body.

Let x

2

be the distance travelled by the second body.

Using s=ut+

2

1

gt

2

,

⇒−x

1

=0×n−

2

1

gn

2

⇒x

1

=

2

1

gn

2

,

⇒−x

2

=0×(n−N)−

2

1

g(n−N)

2

⇒x

2

=

2

1

g(n−N)

2

,

Given x

1

−x

2

=1=

2

1

gn

2

−

2

1

g(n−N)

2

⇒n=

gN

1

+

2

N