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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 it take to reach there?

Given, initial velocity = 5 m/s
Final velocity = 0 m/s
{since the stone will stop at maximum height and then again starts falling down}

$$a = 10 \ m/s^2$$ in the direction opposite to the trajectory of the stone $$= -10 \ m/s^2$$

By the third motion equation,

$$\therefore v^2 - u^2 = 2as$$
$$\Rightarrow s = 1.25 m$$

$$\therefore$$ The distance travelled by the stone = 1.25 meters

Now, using the first motion equation, we get

$$v = u + at$$ $$t = 0.5 s$$

Therefore, time taken by the stone to reach a maximum height = 0.5 seconds