If we drop a stone from a balloon rising at $4\text{m/s}$ , magnitude of initial velocity of stone is taken to be $4$ but why it is so? Please explain by giving more examples.


I think its because the stone is initially travelling along with the baloon with a velocity of 4m/s(with respect to the ground) in the upward direction.So the initial velocity of the stone with respect to the ground is 4m/s.While trying to get off a running bus, you should run along the bus for a small distance.If you try to stand firm (without running), you will fall down.This is because your initial velocity with respect to the ground is not zero.So the ground will try to decelerate you by friction(try to topple you).To prevent such high deceleration, you should run along and gradually come to rest.

  • $\begingroup$ run along the bus in the direction opposite to the direction of the bus $\endgroup$ – PhysicsDave Sep 15 '18 at 14:08
  • $\begingroup$ No.In the same direction as the bus moves(w.r.t ground).Running in the opposite direction will make things worse. $\endgroup$ – Mohan Sep 15 '18 at 14:42
  • $\begingroup$ Are we on the bus trying to get off? In the bus is 10km/hr and you run on the bus 10km/hr is the same direction, total velocity to the ground is 20! $\endgroup$ – PhysicsDave Sep 15 '18 at 16:32
  • $\begingroup$ This is similar to the case of a threadmill.The only difference is that in case of threadmill, the surface wiil be in motion and here you(on the bus) will be initialy in motion.The relative mostion is the same(the threadmill moving back is equivalent to your motion forward).If you are initialy standing over the edge(on the stationary part) of the thread mill and suddenly jump on the moving thread mill, to gain balance you should run in the forward direction.If you stand or run backward, you will fall.Because after jumping, the threadmill will pull the bottom of your leg backward. $\endgroup$ – Mohan Sep 15 '18 at 16:46
  • $\begingroup$ If you want balance, your legs and centre of mass should be in line.So run along the centre of mass(in the ddirection of initial motion of centre of mass) get balance.and slowly come to rest. $\endgroup$ – Mohan Sep 15 '18 at 16:48

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