A projectile is thrown upward so that it's distance (in feet) above ground after t seconds is given by h(t)=-10t^2+440t. What is its maximum height?
ewan ko
To find the maximum height of the projectile, we need to determine the vertex of the quadratic function h(t) = -10t^2 + 440t.
Step 1: Recall that the vertex of a quadratic function is given by the formula (-b/2a, f(-b/2a)), where a, b, and c are the coefficients of the quadratic function ax^2 + bx + c.
Step 2: Identify the values of a, b, and c in the quadratic function -10t^2 + 440t.
In this case, a = -10, b = 440, and c = 0.
Step 3: Substitute the values of a, b, and c into the formula for the x-coordinate of the vertex: x = -b/2a.
Plugging in the values, we get:
x = -440/(2(-10))
x = -440 / (-20)
x = 22
Step 4: Substitute the value of x into the quadratic function to find the y-coordinate of the vertex: f(-b/2a).
Plugging in the value, we get:
f(-b/2a) = -10(22)^2 + 440(22)
f(-b/2a) = -10(484) + 9680
f(-b/2a) = -4840 + 9680
f(-b/2a) = 4840
Therefore, the maximum height of the projectile is 4840 feet.
To find the maximum height of the projectile, we need to determine the vertex of the parabolic function h(t) = -10t^2 + 440t.
The vertex of a parabola in the form y = ax^2 + bx + c is given by the formula:
Vertex = (-b/2a, f(-b/2a))
In this case, a = -10, b = 440, and c = 0 because there is no constant term in the equation.
Using the formula, we can find the value of t at the vertex:
t = -b/2a
t = -440 / (2 * -10)
t = -440 / -20
t = 22
Now, we can substitute this value of t back into the equation to find the maximum height:
h(t) = -10t^2 + 440t
h(22) = -10(22)^2 + 440(22)
h(22) = -10(484) + 9680
h(22) = -4840 + 9680
h(22) = 4840
Therefore, the maximum height of the projectile is 4840 feet.
This is modelled by a downwards opening parabola.
h(t) = -10t^2 + 440t
= -10t(t - 44)
so the x-intercepts are 0 and 44, which means that the vertex lies half way between them
i.e. x = 22
find h(22)
This worked nicely since there was no constant.