
A ball dropped freely takes 0.2 s to travel the last 6 m distance before hitting the ground. Total time of fall is (g = $10m{{s}^{-2}}$):
(A) 2.9 s
(B) 3.1 s
(C) 2.7 s
(D) 0.2 s
Answer
224.4k+ views
Hint: We should know that velocity is defined as the rate change of displacement per unit time. Speed in a specific direction is also known as velocity. Velocity is equal to displacement divided by time. Speed, being a scalar quantity, is the rate at which an object covers distance. The average speed is the distance which is a scalar quantity per time ratio. On the other hand, velocity is a vector quantity; it is direction-aware. An object which moves in the negative direction has a negative velocity. If the object is slowing down then its acceleration vector is directed in the opposite direction as its motion in this case. Based on this we have to solve this question.
Complete step by step answer
Let T be the total time taken in second
S be the total distance travelled in meter
V is the velocity of particle in $\mathrm{m} / \mathrm{sec}$
So, in this question the condition is of free fall so acceleration of the particle will remain constant throughout the process.
So, for time we can write as:
$\mathrm{T}=\mathrm{T}_{1}+\mathrm{T}_{2}$
So, for the distance we can write as:
$S=S_{1}+S_{2}$
Now it is given that: $\mathrm{T}_{1}=0.2 \mathrm{s}$ and $\mathrm{S}_{1}=6 \mathrm{m}$
Thus, we can write that,
$6=u(0.2)+\dfrac{1}{2} 10(0.2)^{2}$
Thus, we get $u=29 m / s$
Now using:
$\mathrm{u}=0+\mathrm{gt}$ (as the ball is dropped with initial velocity as zero)
So, we can find the time t as:
$t=\dfrac{29}{10}=2.9s$
Thus, we get the total time as $2.9+0.2=3.1s$.
Hence, the total time for the fall is 3.1 s. So, the correct option is option B. $2.9+0.2=3.1s$
Note: We should know that if an object's speed or velocity is increasing at a constant rate then we say it has uniform acceleration. The rate of acceleration is constant. If a car speeds up then slows down then speeds up it doesn't have uniform acceleration. The instantaneous acceleration, or simply acceleration, is defined as the limit of the average acceleration when the interval of time considered approaches 0. It is also defined in a similar manner as the derivative of velocity with respect to time. If an object begins acceleration from rest or a standstill, its initial time is 0. If we get a negative value for acceleration, it means the object is slowing down. The acceleration of an object is its change in velocity over an increment of time. This can mean a change in the object's speed or direction. Average acceleration is the change of velocity over a period of time. Constant or uniform acceleration is when the velocity changes the same amount in every equal time period.
Complete step by step answer
Let T be the total time taken in second
S be the total distance travelled in meter
V is the velocity of particle in $\mathrm{m} / \mathrm{sec}$
So, in this question the condition is of free fall so acceleration of the particle will remain constant throughout the process.
So, for time we can write as:
$\mathrm{T}=\mathrm{T}_{1}+\mathrm{T}_{2}$
So, for the distance we can write as:
$S=S_{1}+S_{2}$
Now it is given that: $\mathrm{T}_{1}=0.2 \mathrm{s}$ and $\mathrm{S}_{1}=6 \mathrm{m}$
Thus, we can write that,
$6=u(0.2)+\dfrac{1}{2} 10(0.2)^{2}$
Thus, we get $u=29 m / s$
Now using:
$\mathrm{u}=0+\mathrm{gt}$ (as the ball is dropped with initial velocity as zero)
So, we can find the time t as:
$t=\dfrac{29}{10}=2.9s$
Thus, we get the total time as $2.9+0.2=3.1s$.
Hence, the total time for the fall is 3.1 s. So, the correct option is option B. $2.9+0.2=3.1s$
Note: We should know that if an object's speed or velocity is increasing at a constant rate then we say it has uniform acceleration. The rate of acceleration is constant. If a car speeds up then slows down then speeds up it doesn't have uniform acceleration. The instantaneous acceleration, or simply acceleration, is defined as the limit of the average acceleration when the interval of time considered approaches 0. It is also defined in a similar manner as the derivative of velocity with respect to time. If an object begins acceleration from rest or a standstill, its initial time is 0. If we get a negative value for acceleration, it means the object is slowing down. The acceleration of an object is its change in velocity over an increment of time. This can mean a change in the object's speed or direction. Average acceleration is the change of velocity over a period of time. Constant or uniform acceleration is when the velocity changes the same amount in every equal time period.
Recently Updated Pages
JEE Main 2026 Session 1 Correction Window Started: Check Dates, Edit Link & Fees

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Isoelectronic Definition in Chemistry: Meaning, Examples & Trends

Ionisation Energy and Ionisation Potential Explained

Iodoform Reactions - Important Concepts and Tips for JEE

Introduction to Dimensions: Understanding the Basics

Trending doubts
JEE Main 2026: City Intimation Slip and Exam Dates Released, Application Form Closed, Syllabus & Eligibility

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

Understanding the Angle of Deviation in a Prism

How to Convert a Galvanometer into an Ammeter or Voltmeter

Hybridisation in Chemistry – Concept, Types & Applications

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

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

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

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

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

