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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 ?

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When the stone is thrown vertically upwards, then the velocity of stone goes on decreasing because of force of gravity of earth acting on it in the downward direction. So, the acceleration produced in the stone is negative and hence it is to be written with a minus sign (as, -10 m `s^(-2)`).
Now, `" "` Initial velocity of stone, `u` = 5 m `s^(-1)`
`" "` Final velocity of stone, `v` = 0 `" " ` (It stops at the top)
`" "` Acceleration, `a= -10" m "s^(-2)`
And, `" "` Distance travelled, s = ? `" "` (To be calculated)
`" "` (or Heifht attained)
Now, `" " v^(2) = u^(2) + 2 as`
So, `" "(0)^(2) = (5)^(2) +2 xx (-10)xxs`
`" "0=25 -20 s `
`" "` 20 s = 25
`" " s= (25)/(20)`
`" " s = 1.25 ` m
Thus, the height attained by the stone will be 1.25 metre.
Let us find out the time now. We know that :
`" " v = u+at`
So, `" " 0= 5+(-10)xxt`
`" "0= 5- 10t`
`" " 10 t = 5 `
`" "t =(5)/(10)`
`" " t = 0.5 s `
Thus, the time taken by the stone to reach at the top will be 0.5 second.

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