A stone is thrown vertically upwards with a velocity of 5 m/s . If the acceleration of the stone during the motion is 10 m/s² , then the maximum height attained by the stone is __________ m.

The maximum height attained by the stone is 1.25 m.

Initial velocity, u = 5 m/s
Acceleration, a = - 10 m/s² (The negative sign indicates that the motion of the object is against gravity)
Final velocity at the maximum height, v = 0 m/s

From the third equation of motion,

$2\mathrm{as} =\mathrm{ }{\mathrm{v}}^2\mathrm{ }-\mathrm{ }{\mathrm{u}}^2$
$\Rightarrow$$\mathrm{s}=\frac{{\mathrm{v}}^2\mathrm{ }-\mathrm{ }{\mathrm{u}}^2}{2\mathrm{a}}$
$\Rightarrow$ $\mathrm{s} =\frac{0^2\mathrm{–}{5}^2}{2\times (-10)}\mathrm{ }$
$\Rightarrow$$\mathrm{s}=\frac{-25}{-20}$
$\Rightarrow$$\mathrm{s} =\mathrm{ }1.25 \mathrm{m}$
Thus, the maximum height attained by the stone is 1.25 m.