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Homework Help: Time rockets remains in air

  1. May 1, 2005 #1
    hey, i have a simple question...
    a rocket was launched 33degrees above the horizontal at a speed of 760m/s. it accelerates 15m/ss until it burns out after 10sec. how many sec did it say in the air?

    here s what i did:
    vertical speed = 760sin33 = 413.9 m/s
    horizontal velocity = 637.4 m/s

    so from the vertucal speed, i calculated the time that stays in the air, which is 413.9/9.8= 42.2 sec

    that s the time that took the ricket to go up, so i multiplied by 2, and finally, the time that stayed in the air is 84.4
    BUT, i didn t use the acceleration that happened during 10sec !!!! is what i did above right??
    thank you and i appreciate ur help! :)

  2. jcsd
  3. May 1, 2005 #2
    Calculate the speed (you will only need the vertical component) the rocket gains, in which altitude is the rocket after the acceleration? Then calculate how high the rocket will go with that speed (and the time it takes to go to the highest point). And finally calculate the time it takes the rocket to fall back to Earth. And add the 15s of acceleration.

    Yes, you will have to use the acceleration that's given.
  4. May 1, 2005 #3
    okay...so here is what i did now
    the rocket is goin up by 413.93m/s and an acceleration of 15m/ss during 10sec... the distance that the rocket traveled is 4889.3m and when the rocket burned out after 10sec, it was goin by 564m/s
    no, the rocket is goin 564m/s under the acceleration of -9.8m/ss until it stops...it s gonna take 57.6sec for the rocket to stop, and it traveled 16229.4 m

    now for the rocket to fall back down from a distance of 21118.7 m(16229.4+4889.3=21118.7), it took it 65.7sec

    and finally, 65.7 (time to fall down) + 10 (time where the rocket was accelerating) + 57.6 (when the rocket burned out and was still accelerating until it stops)= 133.3sec

    so 133.3sec is my answer!!! anyone agree?
  5. May 1, 2005 #4
    by the way, if i wanna calculate the momentum of the rocket for example after 5sec, do i use the vertical or horizontal velocity??? (p=mv) and pliiiiz how do i calculate the net force applied to the rocket??
    Last edited: May 1, 2005
  6. May 1, 2005 #5
    I think your calculations are correct but I haven't checked the numerical values.
    If you want to calculate the momentum, yes you can calculate the y- and x-components of the momentum but you can also calculate the momentum going to direction 33 degrees. Momentum has always a direction. You actually can deduce the force if we the air drag is negligible. Just calculate the energy in the start (there's only kinetic energy if we put the zero point of potential energy to 0 m) and after that the kinetic + potential energy right after the acceleration. Then subtract and you'll find the work done by the force. Well, you know the distance s and acceleration a and you'll be able to calculate the mass (and the force of course) with ma*s = E.
  7. May 1, 2005 #6
    thank you!
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