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.

1. isolate the variable, in this case, d. subtract 35.5 on both sides of the equation.2. 52 subtracted by 35.5 is 16.53. therefore your answer would be d= 16.5this is the basics of pre algebra, isolating variable through adding, subtracting, multiplying, and dividing, if you have any other questions to ask just comment!