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Thread: Maximum height of a rocket and how long does this take?

  1. #1
    New Member
    Join Date
    Oct 2005

    Maximum height of a rocket and how long does this take?

    Please can you help with this one?

    A rocket is modelled by a particle which moves along a vertical line. From launch, the rocket rises until its motor cuts out after 18 seconds. At this time it has reached a height of 540 metres above the launch pad and attained an upward velocity of [tex]\ 60ms^{ - 1} \[/tex]. From this time on, the rocket has constant upward acceleration [tex]\ - 10ms^{ - 2} \[/tex] 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 the 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?

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

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

    For part a) I used the formula [tex]\ v^2 - 2as = v_0 ^2 - 2as_0 \[/tex]

    (where [tex]\ v_0 \[/tex] is the initial velocity at the point of origin and [tex]\ s_0 \[/tex] is the initial distance from the origin)

    Putting [tex]\ v_0 = 60 \[/tex] , [tex]\ s_0 = 540 \[/tex] , [tex]\
    a = - 10 \[/tex] and v=0 (max. height is reached when velocity is zero), I got an answer of 720m, but am not sure if this is correct?

    I have got completely stuck on the other two parts. I want to use the formula [tex]\s = \frac{1}{2}at^2 + v_0 t + s_0 \[/tex] but am unsure what values to substitute to find the relevant times. I am quite confused.

    Please can you help?

  2. #2
    Senior Member skeeter's Avatar
    Join Date
    Dec 2005
    Fort Worth, TX
    h<sub>max</sub> will occur 6 seconds after engine cut-off.

    [tex]\L h_{max} = 540 + \int_0^6 60 - 10t dt[/tex]

    to calculate what time the rocket will hit the ground after engine cut-off, solve the quadratic for position ...

    [tex]\L h = h_o + v_o t - \frac{1}{2}gt^2[/tex]

    [tex]\L 0 = 540 + 60t - 5t^2[/tex]


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