What is the average velocity of atoms in 1.00 mol argon at 275k? for m use 0.0399 kg

414.0 m/s

From kinetic theory for an ideal gas ..  

Mean KE of a particle;  ½mv² = 3kT/2

For 1 mol of particles (L); Total KE = L½mv² = 3RT/2

(Lm)½v² = 3RT/2

Lm = molar mass 0.0399kg, R= molar gas constant 8.30 J/(K.mol),  

T = Kelvin temp 275K  

v² = 3RT / Lm  

v² = 3 x 8.30 x 275 / 0.0399

    = 1.716^5

v(rms) = √1.716^5

v(rms) = 414.0 m/s

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be

less than one you can't create energy