Question

4.5.36

Answer 1

Answer:

Step-by-step explanation:

Given function is h(t) = -16t² + 1500

a). For h(t) = 1000 feet,

   1000 = -16t² + 1500

   1000 - 1500 = -16t² + 1500 - 1500

   -500 = -16t²

    t² = $\frac{500}{16}$

    t = $\sqrt{31.25}$

    t = 5.59 sec

b). For h(t) = 940 feet,

   940 = -16t² + 1500

   940 - 1500 = -16t² + 1500 - 1500

   -16t² = -560

   t² = $\frac{-560}{-16}$

   t = $\sqrt{35}$

   t = 5.92 sec

c). For domain and range of the function,

  When the jumper comes down to the ground,

   h = 0

   0 =-16t² + 1500

   t² = $\frac{1500}{16}$

   t = $\sqrt{93.75}$

   t = 9.68 sec

  Since, x-values on the graph vary from x = 0 to x = 9.68,

  Domain : [0, 9.68]

  Vertex of the quadratic function: (0, 1500)

  Since, coefficient of the highest degree term is negative, parabola will open downwards.

  Therefore, y-values of the function will vary in the interval y = 0 to y = 1500

  Range: [0, 1500]