Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
+1 vote
165k views
in JEE by (135 points)

A ball is dropped from the top of a building. The ball takes 0.5s to fall past the 3m height of a window some distance from the top of the building. If the speed of the ball at the top and at the bottom of the window are VT and VB respectively then 

PLEASE GIVE DETAILED SOLUTION OF THE D3QUESTION IN THE ATTATCHMENT

by (56.1k points)
your question is not visible please upload your image again
by (10 points)
+2
Please tell me name of this book

Please log in or register to answer this question.

2 Answers

+1 vote
by (10.9k points)
edited by

A ball is dropped from the top of a building. The ball takes 0.5s to fall past the 3m height of a window som e distance from top of the building If the speeds are of the ball at top and bottom of the window are VT and VB respectively then (g=9.8m/s^2

As acceleration due to gravity  g=9.8m/s^2

We can write VB=VT+ gt

=> VB- VT = 9.8x0.5 =5=4.9m/s .....(1)

VB^2-VT^2=2*g*3=2*9.8*3.....(2)

Dividing (2) by(1) we get

VB+VT =(2*9.8*3)/(4.9)=12m/s

Hence correct option is (A)

+1 vote
by (63.2k points)

We have given, the speed of the ball at the top of the window = VT and 

and the speed of the ball at the bottom of the window = VB

The acceleration we know is ‘g’, g = 9.8 m/s 

The me taken to cross the window, t =0.5 s and the length of the window is S = 3 m. 

We have, V2-U= 2g(3) 

also, V = U+g(0.5) i.e VB-VT = 0.5g 

(VB+VT)(VB-VT) = 6g 

(VB+VT)(0.5g) = 6g 

VB+V= 12 m/s,

VT-V= -4.9 m/s

So option (A) is correct i.e VB+V= 12 m/s

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...