Question

An object is dropped from a height of 100 m . its velocity at the moment it touches the ground
its answer is : 140 m/sec​

Answer 1

Given:

An object is dropped from a height of 100 m . its velocity at the moment it touches the ground.

To find:

Final velocity upon reaching the ground.

Calculation:

Since the object has been dropped hence its initial velocity is zero. Also , the acceleration of the object will be considered constant and equal to gravitational acceleration at the Earth surface.

Now , applying Equation of Kinematics:

$\therefore \: {v}^{2} = {u}^{2} + 2gh$

$= > {v}^{2} = {(0)}^{2} + 2gh$

$= > {v}^{2} = 2gh$

$= > {v}^{2} = 2 \times 10 \times 100$

$= > {v}^{2} = 2000$

$= > v = \sqrt{2000}$

$= > v = 44.72 \: m {s}^{ - 1}$

So, final answer is:

$\boxed{ \bf{v = 44.72 \: m {s}^{ - 1} }}$

Answer 2

Explanation:

Given:

An object is dropped from a height of 100 m . its velocity at the moment it touches the ground.

To find:

Final velocity upon reaching the ground.

Calculation:

Since the object has been dropped hence its initial velocity is zero. Also , the acceleration of the object will be considered constant and equal to gravitational acceleration at the Earth surface.

Now , applying Equation of Kinematics:

\therefore \: {v}^{2} = {u}^{2} + 2gh∴v

2

=u

2

+2gh

= > {v}^{2} = {(0)}^{2} + 2gh=>v

2

=(0)

2

+2gh

= > {v}^{2} = 2gh=>v

2

=2gh

= > {v}^{2} = 2 \times 10 \times 100=>v

2

=2×10×100

= > {v}^{2} = 2000=>v

2

=2000

= > v = \sqrt{2000}=>v=

2000

= > v = 44.72 \: m {s}^{ - 1}=>v=44.72ms

−1

So, final answer is: