# 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

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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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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

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