LIVE Course for free

Rated by 1 million+ students
Get app now
0 votes
in Physics by (66.1k points)
closed by
A stone is thrown in a vertically upward direction with a velocity of `5 m//s`. If the acceleration of the stone during its motion is `10 m//s^(2)` in the downward direction, what will be the height attained by the stone and how much time will take to reach there ?

1 Answer

0 votes
by (68.4k points)
selected by
Best answer
Given initial velocity of stone, u = 5 m `s^(-1)`
Downward of negative Acceleration, a = 10 m `s^(2)`
We know that 2 as = `v^(2)-u^(2)`
Therefore, Height attained by the stone, `s=(0^(2))/(5^(2))xx(-10) m`
=`(-25)/(-20) m `
= 1.25 m
Also we know that final velocity, v = u + at
or, Time, `t=(v-u)/(a)`
Therefore, Time, t taken by stone to attain the height, s `(0-5)/(-10s)`
= 0.5 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.