Answer
353.4k+ views
Hint: In this question, we will use the concept of Newton's equation of motion. We will apply the third equation of motion formula for the constant acceleration and substitute the given values to obtain the final velocity of the stone.
Complete step by step solution
The equations of motion are defined as those which are applied to a particle moving linearly, in three dimensions in a straight line with constant acceleration. By constant acceleration we mean that the velocity is changing linearly with time. In mechanics, acceleration is the rate of change of the velocity of an object with respect to time. Accelerations are vector quantities. The orientation of an object's acceleration is given by the orientation of the net force acting on that object.
Now here in this case we have a constant acceleration acting on the body and that is the acceleration due to gravity. $ g = 9.8m/{\sec ^2} $
We will Use the 3rd equation of motion,
$ \Rightarrow {v^2} = {u^2} + 2as $
Where, $ u $ is the initial velocity of the object
$ v $ is the final velocity of the object before it just touches the ground
$ a $ is nothing but the value of $ g $
$ s $ is the distance covered by the particle from top to bottom of the tower.
Here, we have given that,
$ \Rightarrow s = 19.6\;{\text{m}} $
$ \Rightarrow u = 0\;{\text{m/s}} $
Putting these values in the Equation of Motion we get,
$ \Rightarrow {v^2} = 0 + 2 \times \left( {9.81} \right) \times \left( {19.6} \right) $
We will simplify the above expression as,
$ \Rightarrow v = \sqrt {384.16} $
After simplification we will get,
$ \therefore v = 19.6\;{\text{m/s}} $
Therefore, the final velocity of the particle is $ v = 19.6m/\sec $ .
Note
The acceleration was constant that’s why we could use the equation directly. If the acceleration would have been variable, then we would have to integrate that variable function with respect to time and the usage of a different equation of motion would have been there. The other equations of motion are-
$ \Rightarrow v = u + at $
$ \Rightarrow s = ut + \frac{1}{2}a{t^2} $ .
Complete step by step solution
The equations of motion are defined as those which are applied to a particle moving linearly, in three dimensions in a straight line with constant acceleration. By constant acceleration we mean that the velocity is changing linearly with time. In mechanics, acceleration is the rate of change of the velocity of an object with respect to time. Accelerations are vector quantities. The orientation of an object's acceleration is given by the orientation of the net force acting on that object.
Now here in this case we have a constant acceleration acting on the body and that is the acceleration due to gravity. $ g = 9.8m/{\sec ^2} $
We will Use the 3rd equation of motion,
$ \Rightarrow {v^2} = {u^2} + 2as $
Where, $ u $ is the initial velocity of the object
$ v $ is the final velocity of the object before it just touches the ground
$ a $ is nothing but the value of $ g $
$ s $ is the distance covered by the particle from top to bottom of the tower.
Here, we have given that,
$ \Rightarrow s = 19.6\;{\text{m}} $
$ \Rightarrow u = 0\;{\text{m/s}} $
Putting these values in the Equation of Motion we get,
$ \Rightarrow {v^2} = 0 + 2 \times \left( {9.81} \right) \times \left( {19.6} \right) $
We will simplify the above expression as,
$ \Rightarrow v = \sqrt {384.16} $
After simplification we will get,
$ \therefore v = 19.6\;{\text{m/s}} $
Therefore, the final velocity of the particle is $ v = 19.6m/\sec $ .
Note
The acceleration was constant that’s why we could use the equation directly. If the acceleration would have been variable, then we would have to integrate that variable function with respect to time and the usage of a different equation of motion would have been there. The other equations of motion are-
$ \Rightarrow v = u + at $
$ \Rightarrow s = ut + \frac{1}{2}a{t^2} $ .
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)