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
v2 = u2 + 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.