A rocket moves forward with velocity v = 3c/4, with respect to the earth. Inside this rocket, a second, smaller rocket is launched. In the inertial frame of reference where the larger rocket is at rest, the smaller rocket has velocity v = 3c/4. What is the velocity of the smaller rocket with respect to the earth? You may express the answer as a fraction times c, if you wish (but simplify as much as possible).

Explanation:

We shall apply the formula for relative velocity in relativistic mechanics

Vr = v₁ - v₂ / (1 - v₁v₂/c²) , v₁ is velocity of smaller rocket with respect to earth , v₂ is velocity of large rocket with respect to earth .

Given Vr = 3c/4 , v₁ ? , v₂ = 3c/4

3c/4 = (v₁ - 3c/4) / ( 1 - v₁ 3c/4c²)

3c/4 - (9v₁ / 16) = v₁ - 3c/4

v₁ + 9v₁/16 = 3c/2

= 25v₁ / 16 = 3c / 2

v₁ = 16 / 25 x 3c / 2

= 24/25 x c

it should be -8.6 cm hope this out

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