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
0 votes
220 views
in Physics by (82.2k points)
closed by
A balloon rises from rest on the ground with constant acceleration `g`//`8.` A stone is dropped from the balloon when the balloon has risen to a height of (H). Find the time taken by the stone to reach the ground.
A. (a) ` 4 sqrth/g`
B. (b) ` 2 sqrt h/g`
C. (c ) sqrt (2 h) /g`1
D. (d) ` sqrt g/h`

1 Answer

0 votes
by (84.4k points)
selected by
 
Best answer
Correct Answer - B
Velocity acquired by ballon at heitht (h),
` v= sqrt( 2 ah) = = sqrt 92 xx g/8 h) = (sqrt gh)/2`
Taking downward motion of stone released from balloon at height (h) upto surface of ground, we have
` u= - (sqrt gt)/2 , a=g , S=h, t=t`
As ` S= ut^2`
:. ` h= - 9sqrt gh) /2 t + 1/2 gt^2 or ` t^2 - sqrt h/g t- (2 h) /g =0`.
On solving `, t=2 sqrt h//g ` or - sqrth/g`
As negative tiem is not accounted , hence
` t= 2 sqrt h//g`

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

...