Results 1 to 4 of 4

Math Help - Problem

  1. #1
    Junior Member
    Joined
    Jun 2008
    Posts
    48

    Problem

    A rocket is modelled by a particle which moves along a vertical line. From launch, the rocket rises until its motor cuts out after 16 seconds. At this time it has reached a height of 560 metres above the launch pad and attained an upward velocity of 90ms-1. From this time on, the rocket has constant upward acceleration -10ms-2 (due to the effect of gravity alone).
    Choose the s-axis (for the position of the particle that represents the rocket) to point upwards, with origin at the launch pad. Take t = 0 to be the time when the rocket motor cuts out.

    (a) What is the maximum height (above the launch pad) reached by the rocket?

    v-2as = v0-2as0

    where a = -10ms-2, s0= 0 and v0 = 90

    Thus when v= 0, we obtain

    s= v02 / 2a = 90/2x10 = 405

    So the maximum height reached by the rocket is 405 metres above the launch pad



    b) How long (from launch) does the rocket take to reach this maximum height?

    t= v-v0/a

    v= 10t; a= -10; v0= 90

    = 10-90/-10 = 8

    It take 8 seconds for the rocket to reach the maximum height.




    (c) After how long (from launch) does the rocket crash onto the launch pad?


    where a = -10ms-2, s0= 405 and v0 = 90
    s= 1/2 at + v0t + s0
    = -102/2 + 90t + 405
    = -5t2 + 90t + 405 = 0


    Using the quadratic formula which give t= -9 and t= 12.7



    I have difficulties with the last question, I don't think it's correct
    Can someone help me
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by valerie-petit View Post
    A rocket is modelled by a particle which moves along a vertical line. From launch, the rocket rises until its motor cuts out after 16 seconds. At this time it has reached a height of 560 metres above the launch pad and attained an upward velocity of 90ms-1. From this time on, the rocket has constant upward acceleration -10ms-2 (due to the effect of gravity alone).
    Choose the s-axis (for the position of the particle that represents the rocket) to point upwards, with origin at the launch pad. Take t = 0 to be the time when the rocket motor cuts out.

    (a) What is the maximum height (above the launch pad) reached by the rocket?

    v-2as = v0-2as0

    where a = -10ms-2, s0= 0 and v0 = 90

    Thus when v= 0, we obtain

    s= v02 / 2a = 90/2x10 = 405

    So the maximum height reached by the rocket is 405 metres above the launch pad



    b) How long (from launch) does the rocket take to reach this maximum height?

    t= v-v0/a

    v= 10t; a= -10; v0= 90

    = 10-90/-10 = 8

    It take 8 seconds for the rocket to reach the maximum height.




    (c) After how long (from launch) does the rocket crash onto the launch pad?


    where a = -10ms-2, s0= 405 and v0 = 90
    s= 1/2 at + v0t + s0
    = -102/2 + 90t + 405
    = -5t2 + 90t + 405 = 0


    Using the quadratic formula which give t= -9 and t= 12.7



    I have difficulties with the last question, I don't think it's correct
    Can someone help me
    (c) Find time taken from max height to hitting ground (take DOWNWARDS as positive direction):

    t = T = ?
    initial velocity = 0 m/s
    displacement = 405 m
    acceleration = 10 m/s^2 (since downwards is positive).

    Total time = 8 + T ....

    (For checking purposes I get T = 9 seconds).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jun 2008
    Posts
    48
    Do I have to calculate my quadratic equation with initial velocity = 0?

    Sorry, I'm sure it is easy problem, but really I can't see how it work
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jun 2008
    Posts
    48
    Never mind I got it
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum