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 = [tex]\frac{500}{16}[/tex]
  t = [tex]\sqrt{31.25}[/tex]
  t = 5.59 sec
b). For h(t) = 940 feet,
  940 = -16t² + 1500
  940 - 1500 = -16t² + 1500 - 1500
  -16t² = -560
  t² = [tex]\frac{-560}{-16}[/tex]
  t = [tex]\sqrt{35}[/tex]
  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 = [tex]\frac{1500}{16}[/tex]
  t = [tex]\sqrt{93.75}[/tex]
  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] Â
