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cubanafabien1996

  • one year ago

6. A rocket is launched from atop a 68-foot cliff with an initial velocity of 155 ft/s. The height of the rocket above the ground at time t is given by h = –16t2 + 155t + 68. When will the rocket hit the ground after it is launched? Round to the nearest tenth of a second.

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  1. Hero
    • one year ago
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    Hint: Since h represents the height of the rocket, set h = 0, then solve for t.

  2. Hero
    • one year ago
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    So solve 0 = –16t^2 + 155t + 68

  3. cubanafabien1996
    • one year ago
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    okay thank you

  4. Hero
    • one year ago
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    Did you solve it yet? If so, post your steps and let me know what you got.

  5. cubanafabien1996
    • one year ago
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    not yet.. i'm trying too but idk how too

  6. Hero
    • one year ago
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    Where have you done so far?

  7. Hero
    • one year ago
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    Wow, you're still here?

  8. Hero
    • one year ago
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    Basically, that means you don't know how to factor a quadratic equation.

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spraguer (Moderator)
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is replying to Can someone tell me what button the professor is hitting...

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