Initial velocity of a stone, u = 5 ms^{-1}.

Final velocity of a stone v = 0 (at rest)

If the acceleration of the stone during its motion is = -10ms^{-2}.

Let the maximum height be ‘s’

v = u + at

0 = 5 + (-10)t

t = \(\frac{-5}{10}\) = 0.5s

time taken by stone = 0.5s

As per 3rd Law of motion

v^{2} = u^{2} + 2as

(0)^{2} = (5)^{2} + 2(-10)s

∴ S = \(\frac{5^2}{20}=\frac{25}{20}\) = 1.25 m

Height attained by the store = 1.25 m.