Question

A particle is dropped from a height of 100M what is the distance of particle from the ground after 3 seconds

Answer 1

Concept:

The acceleration of freely falling objects due to the force of attraction of gravitation and centrifugal force is called Acceleration due to gravity.

Given:

The height from which the particle is dropped is $100m$.

Find:

The distance of the particle from the ground after $3$ seconds.

Solution:

The particle is dropped from a height of $100m$.

Initially, the particle is at rest, so the initial velocity is $u=0m/s$.

Since, the particle is in free fall the acceleration due to gravity is, $g=9.8m/s^2$.

Using the second equation of motion,

The distance of particles from the ground after $t=3sec$ is:

$s=ut+\frac{1}{2}gt^2$

$s=0+\frac{1}{2}gt^2$

$s=\frac{1}{2}(9.8)(3)^2$

$s=4.9(9)$

$s=44.1m$

The distance of the particle from the ground after $3sec$ is $44.1m$ .

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Answer 2

Answer:

The distance of particle from the ground after 3 seconds when it is dropped from a height of $100m$ is $44.1m$.

Explanation:

• The acceleration of  objects in free fall due to the gravitational pull and centrifugal force is called gravitational acceleration.

• Given:

 The height from which the particle is dropped $=100m$

• To calculate :

The distance of the particle from the ground after $3$ seconds = ?

The particle is dropped from a height of $100m$.

Initially, the particle is at rest, so the initial velocity of particle $u =0m/s$

As,

the particle is in free fall the acceleration due to gravity                $= -gm/s^{2} = -9.8m/s^{2}$

Using the second equation of motion,

The distance of particles from the ground after $3 seconds$  is $s :$

           $s = ut + \frac{1}{2} gt^{2}$

           $s = 0* 3 +\frac{1}{2} (-9.8) 3^{2}$

           $s = 0 -\frac{1}{2} *9.8*9$

          $s = 4.9* 9$

          $s = 44.1m$

Therefore, the distance of particles from the ground after is $44.1m$

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