Question

a 400 kg rocket is set for vertical firing exhaust speed of the gases is 490m/s. the rate at which gas is to be ejected to give an upward acceleration of g to the rocket

Answer 1

Answer:

16 kg/ s

Explanation:

M(g+g)=(m/t)v

400×2×9.8=(m/t)×490

m/t=16kg/s

When rocket to have an acceleration of g the force is M(g+g) not Mg

Answer 2

Given :

Mass of the rocket (m) = 400 Kg

Speed of the gas ejected (u) = 490

Acceleration (g) = 9.8 m/sec²

To Find :

The rate at which gas is ejecting

Solution :

Let the mass of gas ejected in time dt is dm

Now, thrust force on the rocket ($F_t$) = u ($-\frac{dm}{dt}$)

Net force on the rocket ($F_n_e_t$) = $F_t$ - Weight of the rocket

⇒                                   ma       =  u ($-\frac{dm}{dt}$) - mg     (a = acceleration)

⇒                               m(a + g)   =  u ($-\frac{dm}{dt}$)

As its given that rocket is moving upward with an acceleration 'g' which equals to 9.8 m/sec²

⇒         m(g + g)      =   u ($- \frac{dm}{dt}$)

⇒       $- \frac{dm}{dt}$              =   $\frac{m}{u} 2g$

⇒      $-\frac{dm}{dt}$              =  $\frac{2 *9.8 * 400}{490}$

∴       $-\frac{dm}{dt}$              = 16 Kg/sec

Therefore, The rate at which the gas is to be ejected to give an upward acceleration of g to the rocket is 16 Kg/sec.