A stone with a mass m is dropped from an airplane that has a horizontal velocity v at a height h above a lake. If air resistance is neglected, the horizontal distance R from the point on the lake directly below the point of release to the point where the stone strikes the water is given by which formula? R=v(2h/g)2 None of these are correct. R = 2mv sqrt(2h/g) R = v sqrt(2h/g) R=(1/2)gt2

  R = v √(2h / g)

Explanation:

This exercise can be solved using the concepts of science, projectile launching

let's calculate the time it takes to get to the water

           y = y₀ + t - ½ g t²

as the stone is skipped the vertical speed is zero

           y = y₀ - ½ g t²

for y=0

           t = √ (2y₀ / g)

the horizontal distance it covers in this time is

           R = v₀ₓ t

            R = v₀ₓ √(2 y₀ / g)

let's call the horizontal velocity as v and the height is h

            R = v √(2h / g)

3.99*10^-3N/C

Explanation:

Using

Ep= kq/r²

Where r = 0.6mm = 0.6*10^-3m

K= 8.9*10^9 and q= 1.6*10^-19

So = 8.9*10^9 * 1.6*10^-19/0.6*10^-3)²

= 3.99*10^-3N/C