For a projectile launched from the vertically from the surface the earth with initial velocity v0, the velocity of the projectile when at distance r from the center of the earth satisifes the equation:

v^2 = v0^2 + k^2(1/r - 1/R)

Here, k is a positive constant.
Write the differential equation governing the distance r(t)

(dr/dt) = √(v₀² + k²/r - k²/R)

Explanation:

v² = v₀² + k²[(1/r) - (1/R)]

v = velocity of Body launched from the surface of the earth

v = initial velocity of body

k = a constant

r = distance from the centre of the earth

R = radius of the earth

v² = v₀² + k²/r - k²/R

v = √(v₀² + k²/r - k²/R)

But v = dr/dt

(dr/dt) = √(v₀² + k²/r - k²/R)

(dr/√(v₀² + k²/r - k²/R)) = dt

To go one step beyond and integrate the differential equation

∫ (dr/√(v₀² + k²/r - k²/R)) = ∫ dt

Integrating the left hand side from 0 to r and the right hand side from 0 to t

Note, v₀, k and R are all constants

(- 4r²/k²)[√(v₀² + k²/r - k²/R)] = t

√(v₀² + k²/r - k²/R) = (- k² t/4r²)

(v₀² + k²/r - k²/R) = (- k² t/4r²)²

(v₀² + k²/r - k²/R) = (k⁴t²/16r⁴)

(k⁴t²/16r⁴) + (k²/r) = [v₀² - (k²/R)]

k² [(k²t²/16r⁴) + (1/r)] = [v₀² - (k²/R)]

[(k²t²/16r⁴) + (1/r)] = [(v₀²/k²) - (1/R)]

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s

answer and explanation;