algebra

A ball is thrown into the air with an upward velocity of 40 feet per second. Its height, h, in feet after t seconds is given by the function h(t)=-16t^2+40t+10.
a. What is the ball’s maximum height? 35Ft
b. When the ball hits the ground, how many seconds have passed? Show your work

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  1. I assume you do not do calculus so find the vertex of that parabola. That is just like the last one we did :)
    I will race you.

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  2. 16 t^2 - 40 t - 10 = -h

    divide by 16 first to get 1 as coef of t^2
    t^2 - 2.5 t - .625 = -h/16
    t^2 - 2.5 t = -h/16 + .625
    half of 2.5 then square
    get 1.5625, add to both sides
    t^2 - 2.5 t + 1.5625 =-h/16 + 1.5625
    (t-1.25)^2 = -(1/16)(h-25)
    vertex (top of the parabola) at
    h = 25
    t = 1.25 seconds
    now solve for t when h = 0 .
    Use the + answer

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  3. The vertex is (5/4,35)

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  4. Beat ya :)
    anyway part 2
    16 t^2 - 40 t - 10 = 0
    quadratic equation
    a = 16
    b = -40
    c = -10

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  5. we agree about t, but not about h
    CHECK
    16 t^2 - 40 t - 10 = -h
    16(1.25)^2 - 40(1.25) - 10 = ?
    25 - 50 -10
    -35 = -h , you are right
    I forgot to add 10 on the right

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  6. t = 2.73 for ground time
    https://www.mathsisfun.com/quadratic-equation-solver.html

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