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]
