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