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A stone is thrown in a vertically upward direction with a velocity of $5\, m s^{-1}$. 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 it take to reach there ?
$t=1 \,s\;;\;s=1.25 \, m$
$t=0.5 \,s\;;\;s=2.25 \, m$
$t=1 \,s\;;\;s=2.5 \, m$
$t=0.5 \,s\;;\;s=1.25 \, m$
Solution
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, $s =\frac{0^{2}}{5^{2}} \times(-10) \,m$
$=\frac{-25}{-20} \,m$
$=1.25 \, m$
Also we know that final velocity, $v = u +at$
or, Time, $t =\frac{v-u}{a}$
Therefore, Time, $t$ taken by stone to attain the height, $s=\frac{0-5}{- \,10 \,s}$
$=0.5 \,s$
