Aspace vehicle is traveling at 4150 km/h relative to the earth when the exhausted rocket motor (mass 4m) is disengaged and sent backward with a speed of 79 km/h relative to the command module (mass m). what is the speed of the command module relative to earth just after the separation? km/h

4213.2 Km/h

Explanation:

Given:

Initial Speed of space vehicle relative to Earth, u = 4150 km/h

Mass of rocket motor = 4m

speed of the rocket motor relative to command module , v'  = 79 Km/h

Mass of the command module = m

Now,

let the speed of command module relative to earth be 'v'

From the conservation of momentum, we have

( 4m + m ) × u = m × v + (4m × (v - v'))

or

5m × 4150 = mv + 4mv - 4mv'

or

20750 = 5v - ( 4 × 79 )

or

20750 = 5v - 316

or

v = 4213.2 Km/h

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

the body's situation won't change the forces are equal opposit and on the same body