Question

A body is dropped from certain height and another body is dropped from same height after 2 secs. What will be the separation between two bodies 4 secs after the dropping of second body.

Answer 1

Given :-

• A body is dropped from certain height , let h

• After 2 secs another body is dropped from the same height

To find :-

• Separation (distance) between two bodies 4 secs after dropping the second body

Solution :-

Let us assume that Bodies are dropped from enough height so that they do not reach the ground till 6 sec at least

and the mass of both bodies is equal.

Now,

Calculating distance covered by First body , let s₁

initial velocity of first body , u₁ = 0

time taken by first body , t₁ = 2 + 4 = 6 s

acceleration due to gravity , g = 9.8 ms⁻²

Using second equation motion

S = u t + ½ gt²

⇨ s₁ = u₁t₁ + ½ g t₁²

⇨ s₁ = ( 0 ) (6) +  ½ ( 9.8 ) (6)²

⇨ s₁ = ½ ( 9.8) (36)

⇨ s₁ = 176.4   m

Calculating distance covered by second body , let s₂

initial velocity of second body , u₂ = 0

time taken by second body , t₂ = 4 s

acceleration due to gravity , g = 9.8 ms⁻²

Using second equation of motion

S = u t + ½ gt²

⇨ s₂ = u₂t₂ + ½ g t₂²

⇨ s₂ = ( 0 ) (4) +  ½ ( 9.8 ) (4)²

⇨ s₂ = ½ ( 9.8) (16)

⇨ s₂ = 78.4   m

Calculating the separation b/w two bodies 4 sec after dropping second body

distance b/w bodies = s₁ - s₂

distance b/w bodies = 176.4 - 78.4

→ distance b/w bodies = 98 m.

Hence, separation b/w bodies 4 secs after dropping second body is 98 m .