An archer shoots an arrow into the air. The equation for the height of the arrow in metres (h) over time (t) in seconds is given by the equation h=-2t^2+6t+1. a) Find the maximum height of the arrow b) Find the time for the arrow to reach that height c) Find the height for which the arrow was shot d) Find the length of time the arrow was in the air.
Answers
Answer:
Step-by-step explanation:
a). The maximum height of the arrow is y-coordinate of the vertex of the parabola.
The x-coordinate of the vertex of the parabola is (- b / 2a)
t = - 6 / 2(- 2) = 1.5 seconds
The y-coordinate of the vertex of the parabola is
h = -2(1.5)² + 6(1.5) + 1 = 5.5 (meters) ← maximum height.
b). The x-coordinate of the vertex of the parabola is the time in which the arrow reaches the maximum height.
c). The height for which the arrow was shot is h(0) = 1 meter (y-intercept).
d). - 2t² + 6t + 1 = 0
D = (- 6)² - 4(-2)(1) = 44
= (- 6 - √44) / (- 4) = (- 3 - √11) / (- 2) = 3/2 + √11/2 ≈ 3 seconds
< 0 (not a solution for this task)
Answer:
a). The maximum height of the arrow is y-coordinate of the vertex of the parabola.
The x-coordinate of the vertex of the parabola is (- b / 2a)
t = - 6 / 2(- 2) = 1.5 seconds
The y-coordinate of the vertex of the parabola is
h = -2(1.5)² + 6(1.5) + 1 = 5.5 (meters) ← maximum height.
b). The x-coordinate of the vertex of the parabola is the time in which the arrow reaches the maximum height.
c). The height for which the arrow was shot is h(0) = 1 meter (y-intercept).
d). - 2t² + 6t + 1 = 0
D = (- 6)² - 4(-2)(1) = 44
= (- 6 - √44) / (- 4) = (- 3 - √11) / (- 2) = 3/2 + √11/2 ≈ 3 seconds
< 0 (not a solution for this task)
Step-by-step explanation:
Hope it helps you......Army