If an arrow is shot into the air at a speed of 64 feet per second from a platform that is 30' high. The height of the arrow is given by the function h(t)= -16t2 + 64t + 30, where t is time in seconds. What is the maximum height of the arrow?
the maximum lies on the axis of symmetry of the parabola
tmax = -b / 2a = -64 / (2 * -16)
find tmax , plug into the equation to find max height
To find the maximum height of the arrow, we need to determine the vertex of the quadratic function.
The general form of a quadratic function is given by f(t) = at^2 + bt + c, where a, b, and c are constants.
In this case, the height function is h(t) = -16t^2 + 64t + 30.
The vertex of a quadratic function given by f(t) = at^2 + bt + c is at t = -b/2a.
For our height function h(t), a = -16 and b = 64.
Substituting these values into the formula, we have:
t = -64 / (2 * -16)
t = -64 / -32
t = 2
So, the arrow reaches its maximum height at t = 2 seconds.
Now, let's find the height at this time.
Substitute t = 2 into the height function:
h(2) = -16(2)^2 + 64(2) + 30
h(2) = -16(4) + 128 + 30
h(2) = -64 + 128 + 30
h(2) = 94
Therefore, the maximum height of the arrow is 94 feet.
To find the maximum height of the arrow, we need to determine the vertex of the parabolic function h(t) = -16t^2 + 64t + 30. The vertex represents the highest or lowest point on the graph.
The formula for the x-coordinate of the vertex of a standard quadratic function, represented by the equation ax^2 + bx + c, is x = -b / (2a).
In this case, a = -16 and b = 64. By substituting these values into the formula, we can calculate the value of t for the vertex.
t = -64 / (2 * -16)
t = -64 / -32
t = 2
We have found that the x-coordinate of the vertex is 2. To find the corresponding y-coordinate (which represents the maximum height of the arrow), we substitute t = 2 back into the function h(t).
h(2) = -16(2)^2 + 64(2) + 30
h(2) = -16 * 4 + 128 + 30
h(2) = -64 + 128 + 30
h(2) = 94
Therefore, the maximum height of the arrow is 94 feet.