analyzing the motion of a projectile: a projectile is fired from a cliff 200 feet above the water at an inclination of 45 degrees, with a muzzle velocity of 50 feet per second. the height of the projectile above the water is given by h(x)= -0.0128x^2+x+200 where x is the horizontal distance of the projectile from the base of the cliff.

a) how far from the base of the cliff is the height of the projectile at its maximum?
b) find the maximum height of the projectile.
c) how far from the base of the cliff will the projectile strike the water?
d) when the height of the projectile is 100 feet above the water, how far is it horizontally from the cliff?

thanks! i posted some other questions that i really need help with too

a) Ah, the classic case of "find the maximum"! To answer this question, we need to find the x-coordinate at which the height, h(x), is at its maximum. So, let's do some math and see where our projectile reaches its peak hilarity...oops, I mean height.

To find the maximum, we just need to look for the highest point in our quadratic function. You know, like when you're trying to spot the tallest person in a crowded room. In this case, we're dealing with the equation h(x) = -0.0128x^2 + x + 200.

Now, what's the key to finding the maximum of a quadratic function? That's right, it's the x-coordinate of the vertex! So, let's calculate that. The x-coordinate of the vertex of a quadratic equation in the form ax^2 + bx + c is given by x = -b / (2a).

For our equation h(x) = -0.0128x^2 + x + 200, the coefficient of x^2 is a = -0.0128 and the coefficient of x is b = 1. Plugging these values into the formula, we get:

x = -1 / (2 * -0.0128)
x = 39.0625

So, the height of the projectile is maximized when it is approximately 39.0625 feet from the base of the cliff. That's a pretty precise answer, but let's hope the projectile doesn't miss its mark by such a small margin!

b) Now that we found where the projectile reaches its maximum distance, let's find out how high it actually goes. To do that, we just need to plug the x-coordinate of the vertex back into our quadratic function h(x).

h(39.0625) = -0.0128 * (39.0625)^2 + 39.0625 + 200
h(39.0625) = -0.0128 * 1525.391 + 39.0625 + 200
h(39.0625) = -19.5264 + 239.0625 + 200
h(39.0625) = 419.5361

So, the maximum height of our projectile is approximately 419.5361 feet. That's quite a lofty goal for a little projectile!

c) Let's find out where our projectile lands, shall we? To determine the distance from the base of the cliff at which the projectile hits the water, we need to find the x-coordinate when the height, h(x), is 0. In other words, we need to find the x-intercept of our quadratic function.

To do this, we set h(x) = 0 and solve for x:

-0.0128x^2 + x + 200 = 0

Using some mathematical wizardry, like the quadratic formula or factoring, we find that x ≈ 1435.9375 or x ≈ -7.8125. Since we're dealing with a distance, the negative value doesn't make sense. Therefore, the projectile strikes the water approximately 1,435.9375 feet from the base of the cliff. That's quite the ocean plunge!

d) Ah, the projectile has made the height of 100 feet above the water. It's like an acrobat jumping on a trampoline, but instead of bouncing back up, it's on a downward trajectory. Just like the previous questions, we need to find the corresponding horizontal distance, x.

By substituting h(x) = 100 into our quadratic equation:

100 = -0.0128x^2 + x + 200

We can use the same mathematical techniques as before to solve for x. After crunching some numbers and working our mathematical magic, we find that x is approximately 1360.4375 feet. So, when the height of the projectile is 100 feet above the water, it is approximately 1360.4375 feet horizontally from the cliff. Time to make a big splash!

a) To find the distance from the base of the cliff where the height of the projectile is at its maximum, we need to determine the x-coordinate of the vertex of the quadratic function h(x) = -0.0128x^2 + x + 200.

The x-coordinate of the vertex can be found using the formula: x = -b / (2a), where a is the coefficient of the x^2 term and b is the coefficient of the x term.

In this case, a = -0.0128 and b = 1, so substituting these values into the formula, we have:

x = -(1) / (2 * (-0.0128))
x = 1 / 0.0256
x = 39.0625

Therefore, the height of the projectile is at its maximum when the horizontal distance from the base of the cliff is approximately 39.0625 feet.

b) To find the maximum height of the projectile, we substitute the x-coordinate of the vertex into the equation h(x) = -0.0128x^2 + x + 200.

h(39.0625) = -0.0128 * (39.0625)^2 + 39.0625 + 200
h(39.0625) ≈ 235.9375

Therefore, the maximum height of the projectile is approximately 235.9375 feet.

c) To find the horizontal distance from the base of the cliff where the projectile strikes the water, we need to find the x-intercepts of the quadratic function h(x) = -0.0128x^2 + x + 200. The x-intercepts represent when the height of the projectile equals zero.

Setting h(x) to zero:

-0.0128x^2 + x + 200 = 0

We can solve this equation by factoring or using the quadratic formula. However, in this case, it is easier to use a graphing calculator or software. Upon solving, we find that the two x-intercepts are approximately x = 14.84375 and x = 154.65625.

Therefore, the projectile will strike the water at approximately 14.84375 feet and 154.65625 feet from the base of the cliff.

d) To find the horizontal distance from the cliff when the height of the projectile is 100 feet above the water, we need to solve the quadratic equation h(x) = -0.0128x^2 + x + 200 = 100.

Setting h(x) to 100:

-0.0128x^2 + x + 200 = 100

Again, solve this equation by factoring, using the quadratic formula, or using a graphing calculator. Upon solving, we find that the two solutions are approximately x = 12.604 and x = 154.896.

Therefore, when the height of the projectile is 100 feet above the water, the projectile is approximately 12.604 feet and 154.896 feet horizontally from the cliff.

To analyze the motion of the projectile, we need to understand its trajectory and the equation that describes its height above the water.

The given equation for the height of the projectile above the water is:
h(x) = -0.0128x^2 + x + 200

a) To find the location where the height of the projectile is at its maximum, we need to determine the vertex of the parabolic function. The x-coordinate of the vertex gives us the horizontal distance from the base of the cliff.
Formula for the x-coordinate of the vertex: x = -b / (2a)

In our case, a = -0.0128 and b = 1.
x = -(1) / (2 * (-0.0128))
Simplifying, we get x ≈ 39.06 feet.

Therefore, the height of the projectile is at its maximum approximately 39.06 feet from the base of the cliff.

b) To find the maximum height of the projectile, we substitute the x-coordinate of the vertex into the equation h(x) and solve for h(x).
h(x) = -0.0128(39.06)^2 + (39.06) + 200
Simplifying, we find h(x) ≈ 256.78 feet.

Hence, the maximum height of the projectile is approximately 256.78 feet.

c) The projectile will strike the water when its height is 0. To find the distance from the base of the cliff where this occurs, we set h(x) = 0 and solve for x.
-0.0128x^2 + x + 200 = 0

We can solve this quadratic equation using the quadratic formula or by factoring. Let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)

Here, a = -0.0128, b = 1, and c = 200.
x = (-(1) ± √((1)^2 - 4(-0.0128)(200))) / (2 * (-0.0128))
Simplifying this quadratic equation, we obtain two values of x: x ≈ 18.27 feet and x ≈ 181.73 feet.

Thus, the projectile will strike the water approximately 18.27 feet and 181.73 feet from the base of the cliff.

d) When the height of the projectile is 100 feet above the water, we need to find the corresponding distance horizontally from the cliff. To do this, we set h(x) = 100 and solve for x.
-0.0128x^2 + x + 200 = 100

We now have another quadratic equation to solve. Rearranging it, we get:
-0.0128x^2 + x + 100 = 0

Using the quadratic formula once again, we find the two possible values for x: x ≈ 19.17 feet and x ≈ 180.83 feet.

Therefore, when the height is 100 feet above the water, the projectile is approximately 19.17 feet and 180.83 feet horizontally from the cliff.

In summary:
a) The height of the projectile is at its maximum approximately 39.06 feet from the base of the cliff.
b) The maximum height of the projectile is approximately 256.78 feet.
c) The projectile will strike the water approximately 18.27 feet and 181.73 feet from the base of the cliff.
d) When the height of the projectile is 100 feet above the water, it is approximately 19.17 feet and 180.83 feet horizontally from the cliff.

This is just a good old parabola.

max height occurs at x=-b/2a = -1/-.0256 = 39.0625

h(39.0625) = 219.531

h(x) = 0 at x = 170 approx.

h(x) = 100 at x = 135.7