Astone is thrown upward from ground level. with what minimum speed should the stone be thrown so as to reach a height of 81 feet?

Use the kinematics equation:
v1^2-v0^2=2as
since v1=0 (at height of 81 ft), and a=g=-32.2, substitute values:

0-v0^2=2*(-32.2)(81 ft)
Solve for v0
v0=sqrt(2*32.2*81)=72.2 m/s

An appropriate physics formula is

s=s0 + (1/2)at^2 + v0*t.

Here, the initial height of the stone off the ground is s0=0; the acceleration is g (the acceleration due to gravity)=-32.2 ft/(sec^2); and final height of the stone is 81 ft.  Substituting these values into the above equation gives us

81 ft = 0 ft +(1/2)(-32 ft/(sec^2))(t^2) + v0(t).  We must solve this for the initial velocity, v0.  Unfortunately, we don't yet know how long it will take for the stone to reach its max height, so both t and v0 are variables here.

81 ft = -16 ft/(sec^2)(t^2) + v0(ft/sec)(t)

Gravity will slow the upward progress of the stone:  v = v0 - 16 (ft/sec)(t).
Solving this for t:

0 = v0 - 16 (ft/sec)(t) => t = v0/[-16(ft/sec^2).  Substitute this equation for t into the previous expression and then solve the resulting equation for the initial velocity, v0.

Alternatively, use the following formula, which does not involve time, t:

v^2 = v0^2 + 2 a(s).  

Using v=0, v0 unknown, a = g = -32.2 ft/(sec^2), and s = 81 feet, solve this for v0.

The answe may be c