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

Question 43 Exercise 3E

A body falls from the top of a building and reaches the ground 2.5s later. How high is the building? (Take g=9.8m/s^2)

Given: g=9.8m/s^2, t=2.5s

Assume ‘h’ to be the height of the building

We know from the equation of motion,

\begin{array}{l} \mathrm{H}=\mathrm{ut}+1 / 2 \mathrm{gt}^{2} \\ \Rightarrow \mathrm{H}=1 / 2 \mathrm{gt}^{2} \\ \Rightarrow \mathrm{H}=1 / 2(9.8)(2.5)^{2} \\ \Rightarrow 30.6 \mathrm{m} \end{array}

Video transcript
hello everyone welcome to leader learning i am gurpreet your science tutor today's question is a body falls from the top of a building and reaches the ground 2.5 seconds later how much high is the building take ga nine point eight meter per second square so let's start our answer here the value of j is 9.8 meter per second square time is given as 2.5 second and here we will consider the value of u as 0 from the equation of motion that is h is equals to u d plus half gt square you will get the value of h that is your height here the u is 0 so this term will become 0 and we'll put the values in the next one that is half g is nine point eight into t square that is two point five square you will get the answer as thirty point six meters this is your required answer 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!