Question

the acceleration of a free fall on a particular palnet is 8m/s.an object is droppedfrom a hieght and hits the ground after 1.5 s. from what hieght was it dropped

Answer 1

Given : -

The acceleration of a free fall on a particular palnet is 8m/s². An object is dropped from a hieght and hits the ground after 1.5 s.

Required to find : -

• Height from which the ball is dropped ?

Equations used : -

❶ v = u + at

❷ v² - u² = 2as

Here,

• v = Final velocity

• u = Initial velocity

• a = acceleration

• s = displacement

• t = time taken

Solution : -

The acceleration of a free fall on a particular palnet is 8m/s². An object is dropped from a hieght and hits the ground after 1.5 s.

We need to find the height from which the ball is dropped ?

From the given information we can conclude that ;

• Initial velocity of the object ( u ) = 0 m/s

The object is at rest .

• Acceleration due to gravity ( g ) = 8 m/s²

• Time taken ( t ) = 1.5 seconds

Now,

Let's find the final velocity of the object .

Using the equation of motion ;

i.e. v = u + at

➯ v = 0 m/s + 8 m/s² x 1.5 s

➯ v = 0 m/s + 12 m/s²/s

➯ v = 0 m/s + 12 m/s

➯ v = 12 m/s

Hence,

• Final velocity of the object ( v ) = 12 m/s

Now,

Let's find the displacement of the stone .

Using the equation of motion ;

i.e. v² - u² = 2as

➯ ( 12 )² - ( 0 )² = 2 x 8 x s

➯ 144 - 0 = 2 x 8 x s

➯ 144 = 2 x 8 x s

➯ 144 = 16 x s

➯ 144 = 16s

➯ 16s = 144

➯ s = 144/16

➯ s = 9 meters

Hence,

• Displacement of the object ( s ) = 9 meters

Since,

We can take displacement in terms of distance in some cases .

Therefore,

➯ Height from which the object is dropped is 9 meters

Additional Information

Newton's laws of motion

1st law of motion

A body continues to be in the state of rest or motion untill or unless some external force is applied on it

2nd law of motion

The rate of change of momentum is directly proportional to the unbalanced force in the direction of force .

This law also had given a important formula .

That is ;

Force = mass x acceleration

( F = ma )

3rd law of motion

For every action there is an equivalent opposite reaction .

This can be further complicated and can be stated as ;

If a body A applies some force on body B . Then body B also applies some force of body A which is equal in magnitude but opposite in direction .

This is represented as ;

$\sf{F_{AB} = - F_{BA}}$

Answer 2

Answer:

Given : -

The acceleration of a free fall on a particular palnet is 8m/s². An object is dropped from a hieght and hits the ground after 1.5 s.

Required to find : -

Height from which the ball is dropped ?

Equations used : -

❶ v = u + at

❷ v² - u² = 2as

Here,

v = Final velocity

u = Initial velocity

a = acceleration

s = displacement

t = time taken

Solution : -

The acceleration of a free fall on a particular palnet is 8m/s². An object is dropped from a hieght and hits the ground after 1.5 s.

We need to find the height from which the ball is dropped ?

From the given information we can conclude that ;

Initial velocity of the object ( u ) = 0 m/s

The object is at rest .

Acceleration due to gravity ( g ) = 8 m/s²

Time taken ( t ) = 1.5 seconds

Now,

Let's find the final velocity of the object .

Using the equation of motion ;

i.e. v = u + at

➯ v = 0 m/s + 8 m/s² x 1.5 s

➯ v = 0 m/s + 12 m/s²/s

➯ v = 0 m/s + 12 m/s

➯ v = 12 m/s

Hence,

Final velocity of the object ( v ) = 12 m/s

Now,

Let's find the displacement of the stone .

Using the equation of motion ;

i.e. v² - u² = 2as

➯ ( 12 )² - ( 0 )² = 2 x 8 x s

➯ 144 - 0 = 2 x 8 x s

➯ 144 = 2 x 8 x s

➯ 144 = 16 x s

➯ 144 = 16s

➯ 16s = 144

➯ s = 144/16

➯ s = 9 meters

Hence,

Displacement of the object ( s ) = 9 meters

Since,

We can take displacement in terms of distance in some cases .

Therefore,

➯ Height from which the object is dropped is 9 meters

Additional Information

Newton's laws of motion

1st law of motion

A body continues to be in the state of rest or motion untill or unless some external force is applied on it

2nd law of motion

The rate of change of momentum is directly proportional to the unbalanced force in the direction of force .

This law also had given a important formula .

That is ;

Force = mass x acceleration

( F = ma )

3rd law of motion

For every action there is an equivalent opposite reaction .

This can be further complicated and can be stated as ;

If a body A applies some force on body B . Then body B also applies some force of body A which is equal in magnitude but opposite in direction .

This is represented as ;

\sf{F_{AB} = - F_{BA}}F

AB

=−F

BA