Question

Body 1 of mass 2 kg is dropped from a height h and body 2 of mass 6 kg is dropped
from a height 4 h. Ratio of time taken by the bodies 1 and 2 is:
(a) 1:1
(b) 1:2
(c) 2:1
(d) 1 : 3​

Answer 1

Answer:

• Ratio of time taken by bodies = 1 : 2

Explanation:

Given

• Body 1 is of mass 2 kg and is dropped from a height h
• Body 2 is of mass 6 kg and is dropped from a height 4 h

To find

• Ratio of time taken by the bodies 1 and 2

Formula required

• Second equation of motion

     s = u t + 1/2 a t²

[ Where s is distance covered, u is initial velocity, t is time taken, a is acceleration, t is time taken ]

Solution

While the free fall the mass of the body will not affect the time taken by body to cover a distance, unless we do not consider any air resistance during the Motion.

In this case no air resistance is mentioned to be assumed., therefore, masses of bodies 1 and 2 will not affect the time taken by bodies to reach the ground.

For body 1, taking

• Initial velocity = 0
• acceleration due to gravity = a = g
• distance covered = s₁ = h
• let, time taken to reach the ground = t₁

Using second equation of motion

→ s₁ = u t₁ + 1/2 a t₁²

→ ( h ) = ( 0 ) t₁ + (1/2) (g) t₁²

→ h = g t₁²/2

→ g = 2 h / t₁²      _____equation (1)

Now,

For body 2, taking

• Initial velocity, u = 0
• acceleration due to gravity = a = g
• distance covered = s₂ = 4 h
• let, time taken to reach the ground = t₂

Using second equation of motion

→ s₂ = u t₂ + 1/2 a t₂²

→ ( 4 h ) = ( 0 ) t₂ + ( 1/2 ) ( g ) t₂²

→ 4 h = g t₂² / 2

→ g = 8 h / t₂²      ______equation (2)

By equation (1) and (2)

→ 2 h / t₁² = 8 h / t₂²

→ t₁² / t₂² = (2 h) / (8 h)

→ t₁² / t₂² = 1 / 4

→ t₁ / t₂ = 1 / 2

Therefore,

• Ratio of the time taken by the bodies is., t₁ : t₂ = 1 : 2.

Answer 2

$\Large{\boxed{\underline{\overline{\mathfrak{\star \: QuEsTiOn :- \: \star}}}}}$

Body 1 of mass 2 kg is dropped from a height h and body 2 of mass 6 kg is dropped from a height 4 h. Ratio of time taken by the bodies 1 and 2 is :

$\huge\underline{\overline{\mid{\bold{\pink{ANSWER-}}\mid}}}$

1 : 2

$\Large{\underline{\underline{\bf{Given:-}}}}$

❥ Body 1 is of mass 2 kg and is dropped from a height h

❥ Body 2 is of mass 6 kg and is dropped from a height 4 h

$\Large{\underline{\underline{\bf{Find:-}}}}$

Ratio of time taken by bodies 1 and 2

Here we go !

$\Large{\underline{\underline{\bf{SoLuTion:-}}}}$

Using Second equation of motion

 

   s = u t + 1/2 a t²

" s "is distance

" u " is initial velocity

" t " is time taken

" a " acceleration

Now, According to Question

For body 1,

u = 0

acceleration ( gravity)

= a ( g )

Distance = s₁ = h

Time taken to reach the ground = t₁

As we know ,

↠ s₁ = u t₁ + 1/2 a t₁²

❥ ( h ) = ( 0 ) t₁ + (1/2) (g) t₁²

❥ h = g t₁²/2

❥ g = 2 h / t₁²      .........(1)

For body 2,

u = 0

a = g

distance ↠ s₂ = 4 h

Time taken to reach the ground = t₂

Again according to second law of motion

↠ s₂ = u t₂ + 1/2 a t₂²

❥ ( 4 h ) = ( 0 ) t₂ + ( 1/2 ) ( g ) t₂²

❥ 4 h = g t₂² / 2

❥ g = 8 h / t₂²      . .......... (2)

From (1) & (2) we get

❥ 2 h / t₁² = 8 h / t₂²

❥. t₁² / t₂² = (2 h) / (8 h)

❥ t₁² / t₂² = 1 / 4

❥ t₁ / t₂ = 1 / 2

$\mathfrak{\huge{\purple{\underline{\underline{Hence}}}}}$

Ratio of the time taken by the bodies is.

t₁ : t₂ ↠ 1 : 2

Option B is correct.

More to know :-)

_______________

❥ Free fall means that an object is falling freely with no forces acting upon it except gravity, a defined constant, g = -9.8 m/s2. The distance the object falls, or height, h, is 1/2 gravity x the square of the time falling. Velocity is defined as gravity x time.

❥ An object that falls through a vacuum is subjected to only one external force, the gravitational force, expressed as the weight of the object. An object that is moving only because of the action of gravity is said to be free falling and its motion is described by Newton's second law of motion.

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