The stone is thrown upward from height, h = 85 m from the ground.
It reaches the ground in 5 s.
Let the initial velocity be ‘u’.
The displacement is ‘-85 m’ and acceleration is ‘-10 m/s2’ as they are downward so the negative sign comes.
Using,
S = ut + ½ at2
=> -85 = 5u – ½ × 10 × 52
=> u = 8 m/s
At the greatest height above the ground, the stone comes to momentary rest. That means its velocity becomes zero.
Using,
v2 = u2 + 2as
=> 0 = 82 – 2 × 10 × s
=> s = 3.2 m
Thus, the maximum height from the ground is = 3.2 + 85 = 88.2 m
Velocity with which it reaches the ground is,
v = u + at
=> v = 8 – 10 × 5
=> v = -42 m/s [downward]