Mathematics

Question

A baseball player hit a ball with an upward velocity of 46 ft/s. Its height in feet after t seconds is given by the function h= -16t^2+46t+6. What is the maximum height the ball reaches? How long will it take to reach maximum height? How long does it take to hit the ground?

2 Answer

  • Given the function h(t) where t is the time, to find the maximum height reached by the ball, we must find the maximum value of h(t) = -16t² + 46t + 6. Since h(t) is a quadratic function, to find its maximum value, we need to look for the coordinates of its vertex (x, y). To solve for the x-coordinate, we have

    [tex] x = \frac{-b}{2a} [/tex]
    [tex] x = \frac{-46}{2(-16)} = \frac{23}{16} [/tex]

    Thus, it takes t = 23/16 s for the ball to reach the maximum height. To find the maximum height, we need to substitute the value of t into h(t). 

    h = -16(23/16)² + 46(23/16) + 6 
    h = 625/16 ft

    For the ball to reach ground level, then h must be equal to zero. Thus, we have

    0 = -16t² + 46t + 6 
    0 = 4t² - 23t - 3

    Recalling the quadratic formula, we have

    [tex] t = \frac{-b \pm \sqrt{b^{2}-4ac} }{2a} [/tex]

    where a , b, and c are the coefficients of the quadratic equation. For our case, a = 4, b = -23, and c = -3. Thus, we have

    [tex] t = \frac{23 \pm \sqrt{23^{2}-4(4)(-23)} }{2(4)} [/tex]
    [tex] t = \frac{23 \pm \sqrt{897}}{8} [/tex]

    Since time cannot be negative, we disregard the negative root and have t = (23 + √897)/8 s.

    Answer: 
    Maximum time = 23/16 seconds
    Maximum height = 625/16 ft
    Time taken to reach the ground = (23 + √897)/8 seconds
     

  • Answer:

    maximum height: h=39.06 ft

    time for the ball to reach its maximum height: t= 1.4375 s

    time it takes for the ball to touch the ground: [tex]t_{g}  = 2 * 1.4375 = 2.875s[/tex]

    Step-by-step explanation:

    The function h (t) has a maximum value when h´(t) = 0

    h´(t) is the derivative of h with respect to t.

    We find h´(t)) to calculate the time for the ball to reach its maximum height

    h = -16t²+ 46t + 6 Equation (1)

    h´(t) = -32t + 46

    -32t + 46 = 0

    46 = 32t

    t = 46/32 = 1.4375 s

    We replace 1.4375 s in equation (1) to calculate the maximum height:

    h = -16 (1.4375) ² + 46 (1.4375) + 6

    h = 39.06 ft

    The time it takes for the ball to touch the ground ([tex]t_{g}[/tex]) is twice the time it takes to reach the maximum height.

    [tex]t_{g} =2t[/tex]

    [tex]t_{g}  = 2 * 1.4375 = 2.875s[/tex]

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