Question
Can the ball reach a height of 592?
H=-16^2+64t+512
H=-16^2+64t+512
Answers
oobleck
as with any parabola ax^2+bx+c the vertex is at
(-b/2a, c-b^2/4a)
So, using your parabola
-16t^2+64t+512
the vertex is at (2,576)
So, no, it will not reach a height of 592
(-b/2a, c-b^2/4a)
So, using your parabola
-16t^2+64t+512
the vertex is at (2,576)
So, no, it will not reach a height of 592
henry2,
-16t^2 + 64t + 512 = 592.
-16t^2 + 64t - 80 = 0,
h = Xv = -B/2A = -64/-32 = 2.
K = Yv = -16*2^2 + 64*2 - 80 = -16.
V(h, k) = V(2, -16).
K is < 0, So, we have no real solution.
Therefore, the ball cannot reach 592 ft.
-16t^2 + 64t - 80 = 0,
h = Xv = -B/2A = -64/-32 = 2.
K = Yv = -16*2^2 + 64*2 - 80 = -16.
V(h, k) = V(2, -16).
K is < 0, So, we have no real solution.
Therefore, the ball cannot reach 592 ft.