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Math Help - Equation

  1. #1
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    Equation

    Here's an equation that I don't understand;
    1. The formula to find the height of an object 't' seconds after it is thrown upwards is given by h=ut-5t2(squared)
    where u is the initial speed with which the object is thrown upwards.

    a) Write the equation for an object thrown upwards with an initial velocity of u=80
    so, this is h=80t-5t2

    b) At what times is h=0? (Put h=0 in the equation and solve for t)

    It's the 'b' section that I'm having trouble with....

    Thanks
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  2. #2
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    5t^2-80t=0  <=> 5=\frac{80}{t}  What is t then?
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  3. #3
    A riddle wrapped in an enigma
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    Quote Originally Posted by Tren301 View Post
    Here's an equation that I don't understand;
    1. The formula to find the height of an object 't' seconds after it is thrown upwards is given by h=ut-5t2(squared)
    where u is the initial speed with which the object is thrown upwards.

    a) Write the equation for an object thrown upwards with an initial velocity of u=80
    so, this is h=80t-5t2

    b) At what times is h=0? (Put h=0 in the equation and solve for t)

    It's the 'b' section that I'm having trouble with....

    Thanks
    Hi Tren301,

    Set h = 0 and factor the quadratic to find the two times.

    -5t^2+80t=0

    -5t(t-16)=0

    t=0 \:\:and \:\: t=16

    So, the height is 0 when t = 0. That's at the beginning of the toss.

    The height is 0 when t = 16. That's after it hits the ground at the end of the toss.

    I feel like maybe there should've been some constant in the equation, since
    it would be hard to toss an object from ground level.


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