
: A stone is released from the top of the tower. it covers 24.5 m distance in the last second of its journey. The time, for which stone is air is
A. 3 sec
B. 9 sec
C. 1 sec
D. 5 sec
Answer
231.3k+ views
Hint: for these types of questions, we use the Newton’s equation of motion i.e. $s = ut + \dfrac{1}{2}a{t^2}$, here firstly u = 0 then we can find the value of s. after that it is given that in the last second of the journey i.e. t = t-1, so replace the value of t by t-1 in the value of s and then equate it with 24.5 m, we will get the required result.
Step by step solution:
As, here it is given that the stone is released from the top of the tower, it means initial velocity is zero, and let the stone travel the distance s from the top.
Using Newton’s equation of motion is
$s = ut + \dfrac{1}{2}a{t^2}$
As, u = 0 and as stone is falling under the gravity then take a = g, we get
$ \Rightarrow s = \dfrac{1}{2}g{t^2}$…………. (1)
As, $g = 9.8m{s^{ - 2}}$ then
$ \Rightarrow s = 4.9{t^2}$ …………………. (2)
Now, as it is given that the distance travel by the stone in last second i.e. when t = t-1 seconds is 24.5m, therefore we can substitute the t = t-1 in equation (2), we get
$ \Rightarrow s = 4.9{\left( {t - 1} \right)^2}$ …………………. (3)
Subtracting equation (3) from equation (2), and equate it to 24.5 m, we get
$ \Rightarrow 4.9{t^2} - 4.9{\left( {t - 1} \right)^2} = 24.5$
On solving the above equation, we get
$ \Rightarrow {t^2} - {\left( {t - 1} \right)^2} = 5$
$ \Rightarrow \left( {t - t + 1} \right)\left( {t + t - 1} \right) = 5$
$ \Rightarrow 2t - 1 = 5$
$ \Rightarrow t = 3\sec $
Hence, the time for which stone will remain in air is 3 second
Therefore, option A is correct.
Note: In this equation it must be noticed that when the object is released from the top of the tower it means its initial velocity is equal to zero. After substituting the value we will get the equation.
We have three Newton’s equation of motion i.e. $v = u + at$, ${v^2} - {u^2} = 2as$, $s = ut + \dfrac{1}{2}a{t^2}$
Which equation we have to use this will depend upon the conditions given in the question.
Step by step solution:
As, here it is given that the stone is released from the top of the tower, it means initial velocity is zero, and let the stone travel the distance s from the top.
Using Newton’s equation of motion is
$s = ut + \dfrac{1}{2}a{t^2}$
As, u = 0 and as stone is falling under the gravity then take a = g, we get
$ \Rightarrow s = \dfrac{1}{2}g{t^2}$…………. (1)
As, $g = 9.8m{s^{ - 2}}$ then
$ \Rightarrow s = 4.9{t^2}$ …………………. (2)
Now, as it is given that the distance travel by the stone in last second i.e. when t = t-1 seconds is 24.5m, therefore we can substitute the t = t-1 in equation (2), we get
$ \Rightarrow s = 4.9{\left( {t - 1} \right)^2}$ …………………. (3)
Subtracting equation (3) from equation (2), and equate it to 24.5 m, we get
$ \Rightarrow 4.9{t^2} - 4.9{\left( {t - 1} \right)^2} = 24.5$
On solving the above equation, we get
$ \Rightarrow {t^2} - {\left( {t - 1} \right)^2} = 5$
$ \Rightarrow \left( {t - t + 1} \right)\left( {t + t - 1} \right) = 5$
$ \Rightarrow 2t - 1 = 5$
$ \Rightarrow t = 3\sec $
Hence, the time for which stone will remain in air is 3 second
Therefore, option A is correct.
Note: In this equation it must be noticed that when the object is released from the top of the tower it means its initial velocity is equal to zero. After substituting the value we will get the equation.
We have three Newton’s equation of motion i.e. $v = u + at$, ${v^2} - {u^2} = 2as$, $s = ut + \dfrac{1}{2}a{t^2}$
Which equation we have to use this will depend upon the conditions given in the question.
Recently Updated Pages
Dimensions of Pressure in Physics: Formula, Derivation & SI Unit

Circuit Switching vs Packet Switching: Key Differences Explained

JEE General Topics in Chemistry Important Concepts and Tips

JEE Extractive Metallurgy Important Concepts and Tips for Exam Preparation

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Electricity and Magnetism Explained: Key Concepts & Applications

Trending doubts
JEE Main 2026: Admit Card Out, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Laws of Motion Class 11 Physics Chapter 4 CBSE Notes - 2025-26

Waves Class 11 Physics Chapter 14 CBSE Notes - 2025-26

Mechanical Properties of Fluids Class 11 Physics Chapter 9 CBSE Notes - 2025-26

Thermodynamics Class 11 Physics Chapter 11 CBSE Notes - 2025-26

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

