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A projectile is fired from a cliff 100 ft above the water at an inclination of 45 degrees to the horizontal, with a muzzle velocity of 260 ft/second. The height of the projectile above the water is given by h(x)= ((-32x^2)/(260)^2)+x+100, where x is the horizontal distance of the projectile from the base of the cliff. How far from the base of the cliff is the height of the projectile a maximum?
The way you typed it, the equation reduces to
h(x) = (-2/4225)x^2 + x + 100
h'(x) = (-4/4225)x + 1 = 0 for a max height
-4x = -4225
x = 1056.25 ft from base of cliff
evaluate h(1056.25) to get the actual max height.
To find the horizontal distance at which the height of the projectile is a maximum, we need to find the vertex of the quadratic equation h(x) = ((-32x^2)/(260)^2) + x + 100.
The vertex of a quadratic equation in the form ax^2 + bx + c is given by the x-coordinate x = -b / (2a).
In this case, a = -32/ (260)^2 = -32 / 67600 and b = 1. There is no c term in our equation, so c = 0.
Using the formula, we can find the x-coordinate of the vertex:
x = -b / (2a)
= -1 / (2 * (-32/ (260)^2))
= -1 / (-64/ 67600)
= 67600 / 64
= 1056.25
Therefore, the horizontal distance at which the height of the projectile is a maximum is approximately 1056.25 ft.
To find the horizontal distance at which the height of the projectile is a maximum, we need to find the x-value that corresponds to the maximum point of the quadratic function h(x).
The quadratic function h(x) is defined as:
h(x) = (-32x^2)/(260^2) + x + 100
To find the maximum point of the function, we need to find the x-value where the derivative of h(x) is equal to zero.
Step 1: Find the derivative of h(x)
To do this, we differentiate h(x) with respect to x:
h'(x) = d/dx [(-32x^2)/(260^2)] + d/dx [x] + d/dx [100]
h'(x) = (-64x)/(260^2) + 1 + 0
Simplifying, we get:
h'(x) = (-64x)/(260^2) + 1
Step 2: Set h'(x) = 0 and solve for x
To find the x-value where h'(x) is equal to zero, we set the derivative equal to zero and solve for x:
(-64x)/(260^2) + 1 = 0
Multiply both sides by (260^2) to eliminate the denominator:
-64x + (260^2) = 0
-64x = - (260^2)
Divide both sides by -64:
x = (260^2)/64
Simplifying, we get:
x ≈ 1067.19 ft
So, the horizontal distance at which the height of the projectile is a maximum is approximately 1067.19 ft from the base of the cliff.