1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Maximum altitude of a rocket engine

  1. Sep 1, 2010 #1
    1. The problem statement, all variables and given/known data
    A rocket is launched at an angle of 53 above the horizontal with an initial speed of 100 m/s. It moves along its initial line of motion with an acceleration of 30 m/s^2 for 3 s. A this time its engines fail and the rocket proceeds to move as a free body.

    a. Find maximum altitude. =1521.5 m (Correct answer, according to some sources. . .)
    b. Its total time of flight = 36.1 s (Correct answer, according to some sources. . .)
    c. It's range =4045 m (Correct answer, according to some sources. . .)

    I have completed this problem, however my answers do not seem to agree with what the correct answers are supposed to be. The correct answers are noted above. Could someone please point out where I went wrong? I feel like I understand why my process should work, but it's clearly wrong. . .

    3. The attempt at a solution
    a.
    -The engine stops working at this vertical height: dy =100 sin 53 (3) + 1/2 (30) (3^2)=374.590653
    -The rocket's speed after the engines fail is: vf= vi + at = 100 + 30 (3) = 190 m/s
    -When the rocket fails, it is still moving at the original angle. Resolved, the vertical velocity is: viy= 190sin53

    So: vf^2= vi^2 - 2gd ---> d = -vi^2 / (-2 (9.8)) = 1174.757871

    Then, to find the maximum altitude: 1174.757871 + 374.590653 = 1549.35 m

    b.
    -Time with the engine working is given = 3s.

    -Time w/o the engine: vf = vi +at --> 0 = 190sin53 -9.8t --> t= 15.48374968s

    -Time to fall: vf= vi + at --> t= vf-vi/a--> t= (squareroot 2ad )/a --> t= (squareroot 2 * 9.8 * 1549.35)/ 9.8 = 17.78184123 s

    Adding all three values together = 36.26559091 s.


    c.
    There are two different horizontal velocities: 100cos53 and 190cos53. I solve for both:

    For the acceleration period, the horizontal range should be:
    dx= 100cos53 (3) + 1/2 (30) 3^2
    dx= 315.5445069

    For the time after the engines fail:
    dx= 190cos53 (33.26559091) + 0 (there is no horizontal acceleration after the engines fail)

    Added together, the two values give a horizontal range of= 4119.293656 m.


    Thank you in advance! I very much appreciate any help given to identify what I am doing wrong!!
     
  2. jcsd
  3. Sep 1, 2010 #2
    In First Part you have not taken component of rocket's initial acceleration in y direction
    dy =100 sin 53 (3) + 1/2 (30) (3^2)sin53 = 347.406
     
  4. Sep 1, 2010 #3
    In Third Part you have not taken component of rocket's initial acceleration in x direction
    When engine fails at this point use
    equation of trajectory
    y=xtanA-(gx^2)/(2u^2Cos^2A)
    A= 53
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Maximum altitude of a rocket engine
  1. Maximum altitude (Replies: 7)

Loading...