Quadratic equations motion


Can anyone offer any advice on solving the following,

motion of two stage rocket is stated by it height y in metres above the ground since t seconds since launch.

Flies vertically Ist stage burns for 5 seconds and cuts out. 2nd stage rocket is under gravity alone.

modelled by the equations

Stage 1 height y = 10t2

time 0< t <5

stage 2 height y = -5t2 + 150t - 375

time t >5

Can anyone offer some help... I need to work out speed velocity etc,

Thank you

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  1. during the first 5 seconds the height is given by 10 t^2
    So it reaches an altitude of 10(25) = 250 meters before dropping the first stage.
    At that time, 5 seconds, the velocity is given by dy/dt = 20 t = 100 m/s

    Then stage two continues. It shows no sign of being a rocket since it seems to be operating only under gravity with an initial height of 250 meters at t = 5 and an initial velocity of 100 m/s at t = 5
    Its velocity is determined by
    dy/dt = -10 t + 150
    at the maximum height, that velocity will be zero
    0 = -10 t + 150
    or t = 15 seconds at peak
    then the max height will be
    y = -5 (225) + 150 (15) - 375
    y = -1125+2250-375
    y = 750 meters at peak.
    Then when will it hit the ground?
    0 = -5 t^2 + 150 t -375
    t^2 -30 t + 75 = 0
    t=(1/2) [30 +/-sqrt(900-300)]
    t = (1/2)[30 +/- 24.5]
    Well, the minus sign will not do because the thing is still headed with stage 1 up so use the + sign
    t = (1/2)(54.5) = 27.25 seconds to crash

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  2. If you mean
    y = 10t^2 for 0<t<5 s and
    y = -5t^2 +150 t -375 for t>5 (until t = 25 s, when y = 0), then
    the velocity (and speed) are
    20 t for t<5 and
    -10t + 150 for 5<t<25

    The acceleration is 20 during t<5s and -10 during 5<t<25s

    My answers were obtained by differentiation. I assume you are studying calculus.

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  3. Thank you for your help

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  4. My statement that y=0 at t=25 s is incorrect. Damon derived the correct value

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  5. Thank you ..

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  6. Thank you have a good weekend !!

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