A baseball is thrown upward and its height after t seconds can be described by formula h(t)=−16t^2+50t+5. Find the maximum height the ball will reach.
Just FYI, which may come in handy, the y-coordinate of the vertex is
(4ac-b^2)/(4a)
or
c - b^2/(4a)
To find the maximum height the ball will reach, we need to determine the vertex of the parabolic function h(t) = -16t^2 + 50t + 5.
The vertex of a parabola is given by the formula (h, k), where h is the x-coordinate of the vertex and k is the y-coordinate of the vertex. In this case, the x-coordinate represents time, t, and the y-coordinate represents height, h(t).
The x-coordinate of the vertex, h, can be found using the formula h = -b/2a, where a, b, and c are the coefficients in the quadratic equation. From the equation h(t) = -16t^2 + 50t + 5, we can see that a = -16 and b = 50. Substituting these values into the formula, we get h = -50 / (2 * -16) = -50 / -32 = 1.5625.
Now, to find the y-coordinate of the vertex, k, we substitute the value of h back into the original equation h(t) = -16t^2 + 50t + 5. Thus, k = -16(1.5625)^2 + 50(1.5625) + 5 = 36.5625 + 78.125 + 5 = 119.6875.
Therefore, the maximum height the ball will reach is approximately 119.6875 units.
You need the vertex : )
Either factor to find the zeros then find the halfway point between them (sub in the x-value and solve for y) or use x = -b/(2a) to find the x-coordinate of the vertex : )
Your choice...
Given:
h(t) = -16t^2 + 50t + 5.
g = 32ft/s^2
Vo = 50 ft./s
V = Vo + g*t = 0.
50 + (-32)t = 0,
t = 1.56 s.
h = -16*1.56^2 + 50*1.56 + 5 = 44.1 Ft.