 # Selina Solutions Class 9 Physics Solutions for Exercise 3E in Chapter 3 - Chapter 3 Laws Of Motion

A body falls freely under gravity from rest and reaches the ground in time t. Write an expression for the height fallen by the body.

Given: acceleration due to gravity=’g’, initial velocity, u=0, time=t.

Let ‘h’ be the height fallen by the body

h = ut+\frac{1}{2}gt^2

u = 0 m/s, therefore the equation becomes;

h = \frac{1}{2}gt^2

Video transcript
hello everyone welcome to leader learning i'm good freak your science tutor today's question is a body falls freely under the gravity from rest and reaches the ground in time t write an expression for the height fallen by the body so here in this question you have to write the height of the body and how it is expressed so let's start our answer here acceleration due to gravity due to gravity is g initial velocity that is u is equals to 0 because the body is falling from the rest and time is given as t time if we suppose h is the height fallen by the body then it can be expressed as h is equals to u d plus half gt square since the value of u here is 0 your final expression will be h is equals to half gt square this will be the required expression for more such videos please subscribe leader learning and for any doubts drop a comment thank you
Related Questions

Lido

Courses

Teachers

Book a Demo with us

Syllabus

Maths
CBSE
Maths
ICSE
Science
CBSE

Science
ICSE
English
CBSE
English
ICSE
Coding

Terms & Policies

Selina Question Bank

Maths
Physics
Biology

Allied Question Bank

Chemistry
Connect with us on social media!