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A rocket is fired vertically and ascends with a constant vertical acceleration of 20 m/s^2 for 1 minute. It's fuel is then all used and it continues as a free particle. Find the

  1. Maximum height reached by the particle

  2. Total time elapsed from the take off till the rocket strikes the earth.

I cannot understand this question practically. First of all does the acceleration include $g$ in it . Also is it deacceleration or acceleration in this case. If it is acceleration then how can the velocity become zero in 60 sec. Also is the initial velocity zero ? I feel that there should be an initial velocity like in the case of projectile motion. Please guide me in this question.

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  • $\begingroup$ The velocity does not become zero in 60 seconds. Continuing as a free particle means it becomes a projectile. That does not mean its velocity goes to zero $\endgroup$ – Jim Aug 9 '14 at 19:22
  • $\begingroup$ For the first stage acceleration is $20 {\rm nm/s}$ and for the second stage it is $-g$. Now integrate the acceleration to get velocity and height. $\endgroup$ – ja72 Aug 9 '14 at 19:22
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First of all does the acceleration include $g$ in it

The problem can be worded in two different ways:

  1. "The position with respect to the ground changes with constant acceleration $20\ \mathrm{m/s^2}$."
  2. "The passengers feel a constant acceleration of $20\ \mathrm{m/s^2}$."

Since there is no mention of passengers' feelings, I'm inclined to think the problem means interpretation (1). Can you see how (2) would lead to an acceleration with respect to the ground of $20\pm10\ \mathrm{m/s^2}$? (I leave the determination of the sign as an exercise to the reader.)

Also is it deacceleration or acceleration in this case

The word "deceleration" is really only useful if you want to avoid specifying directions explicitly, but want to convey that the direction of acceleration is opposite the current direction of velocity.

Now in the sense of (1) above, I'd imagine the acceleration is upward (rockets don't blast themselves toward the ground too often). And if the rocket is also moving upward at the time, then this is not a deceleration.

If it is acceleration then how can the velocity become zero in 60 sec. Also is the initial velocity zero

There are two parts to the question: when the rocket is powered and when it is not. Both have initial conditions (a height and a velocity).

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  • $\begingroup$ So what I understand is that even after the fuel gets over then also the rocket will move some distance which I will need to take into account right? $\endgroup$ – Jai Mahajan Aug 9 '14 at 19:25
  • $\begingroup$ @user166748 Yes, generally an upward-directed object coasts upward for a while even after its source of thrust has disappeared. $\endgroup$ – user10851 Aug 9 '14 at 19:26
  • $\begingroup$ Ok... Also i didn't understand why deacceleration in this case means that it will crash into the ground. Supposedly the rocket is fired with an initial velocity and then it can have a deacceleration due to gravity and it will still travel upward. $\endgroup$ – Jai Mahajan Aug 9 '14 at 19:31
  • $\begingroup$ @user166748 I'm referring to the powered portion of its flight, where it starts out near the ground not moving. If you're near the ground not going fast and then you decide to accelerate downward, you're playing a dangerous game. $\endgroup$ – user10851 Aug 9 '14 at 19:33
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First of all does the acceleration include g in it

When the acceleration is given like that, all influences (forces) causing the acceleration to have that value have been taken into account. In other words, the specified acceleration is the acceleration, there is no need to add or subtract $g$ from it.

On the other hand, once the engines turn off, the rocket is in free fall and its acceleration is purely the acceleration due to gravity.

is it deacceleration or acceleration in this case

Presumably, the rocket is moving "up" away from the Earth, so if you choose your coordinates such that "up" is the positive direction, then the acceleration in that direction is $a = +20\,\mathrm m/\mathrm s^2$

is the initial velocity zero

Most probably yes since rocket launches in the real world typically take place with the rocket starting from rest on the surface of the Earth.

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  • $\begingroup$ So what I understand is that even after the fuel gets over then also the rocket will move some distance which I will need to take into account right? $\endgroup$ – Jai Mahajan Aug 9 '14 at 19:25

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