?
Lv 4
? asked in Science & MathematicsMathematics · 2 months ago

Math Question Help?

A stunt pilot is testing a new plane. The equation that models his height over time is h(t) = 15t^2 - 195t + 950, where t is time in seconds and h(t) is his height in meters. Determine when the stunt pilot and the test plane are below 500 meters.

I'm completely lost on how to solve this, please help

6 Answers

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  • ?
    Lv 7
    2 months ago
    Favorite Answer

    h (t) = 15t² - 195t + 950

    500 >  15t² - 195t + 950

    15t² - 195t + 950 - 500 < 0

    15t² - 195t + 450 < 0

    15 (t² - 13t + 30) < 0

                  15 (t - 10)(t - 3) < 0

                   when t < 10 and t > 3 .............ANS

    Attachment image
  • 2 months ago

    h(t)=15t^2-195t+950<500

    =>

    t^2-13t+30<0

    =>

    (t-10)(t-3)<0

    =>

    3<t<10

    =>

    The plane are below 500 m

    when it is more than 3 seconds

    & less than 10 seconds after

    starting the flight.

  • 2 months ago

    We need to find the times when the pilot is at 500 metres.

    i.e. when 15t² - 195t + 950 = 500

    so, 15t² - 195t + 450 = 0

    or, t² - 13t + 30 = 0

    Hence, (t - 3)(t - 10) = 0

    so, at t = 3 and again at t = 10

    As the function is a parabola with a ' ∪ ' shape, the pilot is below 500 metres between these times

    i.e. below 500 metres from t = 3 to t = 10 seconds

    Hence, 7 seconds in all

    :)>

  • Ash
    Lv 7
    2 months ago

    h(t) = 15t² - 195t + 950

    We want to find time when h(t) <500

    15t² - 195t + 950 < 500

    15t² - 195t + 450 < 0

    t² - 13t + 30 < 0

    (t-10)(t-3) < 0

    This gives 2 options

    Option-1

    (t-10)>0 and (t-3) < 0, this gives...

    t>10 and t<3

    This is absurd , so ignore

    Option-2

    (t-10)<0 and (t-3) > 0, this gives...

    t<10 and t>3

    That means the test plane is below 500m between 3 seconds to 10 seconds

    Attachment image
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  • Jim
    Lv 7
    2 months ago

    h(t) = 15t^2 - 195t + 950  ((see red graph))

    Find  15t^2 - 195t + 950 < 500, so 15t^2 - 195t + 950 - 500 < 0

    Find instead where it equals 0

    15t^2 - 195t + 450 = 0, solve using Quadratic

    At t between 3 & 10, height will be less than 500m

    Attachment image
  • ted s
    Lv 7
    2 months ago

    set h(t) = 500 and solve for t values...then check the values of h(t) on either side of those t values

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