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Elementary algebra help!!!?

A water rocket is launched upward with an initial velocity of 48ft/sec. Its height h, in feet, after t seconds is given by h=48t - 16t^2. Determine the height of the rocket 2.5 sec after it has been launched.

I need the procedure please, explainn me!!

7 Answers

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  • 1 decade ago
    Favorite Answer

    h = 48 (2.5) - 16 (2.5)^2

    h = 120 - 16(6.25)

    h = 120 - 100

    h = 20 ft

  • 1 decade ago

    h = 48t - 16t^2

    Since t = 2.5 sec, I guess you have just to plug it in the equation.

    h = 48 (2.5) - 16 (2.5)^2

    h = 120 - 16(6.25)

    h = 120 - 100

    h = 20 feet ===> the answer

  • Wyk123
    Lv 4
    1 decade ago

    just substitute 2.5 sec. to the given equation:

    h=48*2.5+16*(2.5)^2

    h= 120 +100

    h=320 ft

  • looikk
    Lv 4
    1 decade ago

    h=48t - 16t^2

    at t = 2.5 sec

    height, h = 48(2.5) - 16(2.5)^2 = 20 ft

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  • Anonymous
    1 decade ago

    Just substitute 2.5 for t every time t occurs.

    h is height.

    h= 48t - 16t^2

    h= 48(2.5) - 16(2.5)(2.5) now calculate it.

  • 1 decade ago

    Just plug in 2.5 seconds for t (since t is time here) so

    48(t) -16(t)^2=h

    this is basically using the energy equation (assuming no non-conservative forces)

  • 1 decade ago

    put 2.5 in for t

    h=(48*2.5)-(16*(2.5*2.5))

    h=120-100

    h=20ft

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