Question.
A stone is thrown in vertically upward direction with a velocity of 5 ms–1. If the acceleration of the stone during its motion is 10 ms–2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?
A stone is thrown in vertically upward direction with a velocity of 5 ms–1. If the acceleration of the stone during its motion is 10 ms–2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?
Solution:
Given, initial velocity, $\mathrm{u}=5 \mathrm{~ms}^{-1}$
Final velocity, $\mathrm{v}=0$
Since, $\mathrm{u}$ is upward \& a is downward, it is a retarded motion. $\therefore \mathrm{a}=-10 \mathrm{~ms}^{-2}$
Height attained by stone, $\mathrm{s}=?$
Time taken to attain height, $\mathrm{t}=?$
(i) Using the relation, $\mathrm{v}=\mathrm{u}+\mathrm{at}$
$0=5+(-10) t$ or
$t=5 / 10=0.5 \mathrm{~s}$
(ii) Using the relation, $v^{2}-u^{2}=2$ as, we have
$s=\frac{v^{2}-u^{2}}{2 a}=\frac{(0)^{2}-(5)^{2}}{2 \times(-10)}=1.25 \mathrm{~m}$
Given, initial velocity, $\mathrm{u}=5 \mathrm{~ms}^{-1}$
Final velocity, $\mathrm{v}=0$
Since, $\mathrm{u}$ is upward \& a is downward, it is a retarded motion. $\therefore \mathrm{a}=-10 \mathrm{~ms}^{-2}$
Height attained by stone, $\mathrm{s}=?$
Time taken to attain height, $\mathrm{t}=?$
(i) Using the relation, $\mathrm{v}=\mathrm{u}+\mathrm{at}$
$0=5+(-10) t$ or
$t=5 / 10=0.5 \mathrm{~s}$
(ii) Using the relation, $v^{2}-u^{2}=2$ as, we have
$s=\frac{v^{2}-u^{2}}{2 a}=\frac{(0)^{2}-(5)^{2}}{2 \times(-10)}=1.25 \mathrm{~m}$