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
86 views
in Physics by (66.1k points)
closed by
A stone is released from the top of a tower of height 19.6m. Calculate its final velocity just before touching the ground.

1 Answer

0 votes
by (68.5k points)
selected by
 
Best answer
Here, Initial velocity u=0
final velocity v=? (to be calculated )
Acceleration due to gravity `g=9.8 m//s^(2)` (Stone comes down )
And Height h=19.6 m
Now ,we know that for a freely falling body :
`v^(2)=u^(2)+2gh `
So , ` v^(2)=(0)^(2)+2xx9.8xx19.6`
`v^(2)=19.6xx19.6`
`v^(2)=(19.6)^(2)`
v=19.6 m/s
Thus the velocity of stone just before hitting the ground will be 19.6 metres per second .

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

...