quadratic equation involving h(t): h(t) = -16t^2 + vt + s

[bhardin]

h(t)= -16t^2+vt+s

The given information: The cannon shoots 10 feet above the ground. The max height of 30 feet occurs 40 feet from the cannon.

I know that s=10, but I don't know how to find (v). I am confused.

 

[Deleted member 4993]

h(t)= -16t^2+vt+s

The given information: The cannon shoots 10 feet above the ground. The max height of 30 feet occurs 40 feet from the cannon.

I know that s=10, but I don't know how to find (v). I am confused.

Above is an equation for parabola.

What is the location of vertex?

 

[stapel]

h(t)= -16t^2+vt+s

The given information: The cannon shoots 10 feet above the ground. The max height of 30 feet occurs 40 feet from the cannon.

I know that s=10, but I don't know how to find (v). I am confused.

What is "s"? Is it perhaps "the initial height"? (Usually, the projectile equation itself is named "s(t)", and "h" is the initial height.)

Since you "know that s = 10", plus that into the equation:

. . . . .\(\displaystyle h(x)\, =\, -16t^2\, +\, vt\, +\, 10\)

You are told that the max value of the height "h" is 30. You've been given a hint of thinking about the vertex. I'm not sure how well that might work, given what we have. However:

You know that the vertex occurs when the height is 30. This means that you have:

. . . . .\(\displaystyle 30\, =\, -16t^2\, +\, vt\, +\, 10\)

. . . . .\(\displaystyle 16t^2\, -\, vt\, +\, 20\, =\, 0\)

You know that this cannonball's path can have only one vertex, so this equation should given you exactly one solution.

. . . . .\(\displaystyle t\, =\, \dfrac{-(-v)\, \pm\, \sqrt{\strut (v)^2\, -\, 4(16)(20)\,}}{2(16)}\)

. . . . . . .\(\displaystyle =\, \dfrac{v\, \pm\, \sqrt{\strut v^2\, -\, 1280\,}}{32}\)

Since this can have only one solution, what then must be true about the expression inside the square root?

Use this fact to find the value of the initial velocity "v", remembering that the cannonball was moving upward, which is a "positive" direction.

 

[bhardin]

Thank you. The equation makes some sense to me now.

Thanks

What is "s"? Is it perhaps "the initial height"? (Usually, the projectile equation itself is named "s(t)", and "h" is the initial height.)

Since you "know that s = 10", plus that into the equation:

. . . . .\(\displaystyle h(x)\, =\, -16t^2\, +\, vt\, +\, 10\)

You are told that the max value of the height "h" is 30. You've been given a hint of thinking about the vertex. I'm not sure how well that might work, given what we have. However:

You know that the vertex occurs when the height is 30. This means that you have:

. . . . .\(\displaystyle 30\, =\, -16t^2\, +\, vt\, +\, 10\)

. . . . .\(\displaystyle 16t^2\, -\, vt\, +\, 20\, =\, 0\)

You know that this cannonball's path can have only one vertex, so this equation should given you exactly one solution.

. . . . .\(\displaystyle t\, =\, \dfrac{-(-v)\, \pm\, \sqrt{\strut (v)^2\, -\, 4(16)(20)\,}}{2(16)}\)

. . . . . . .\(\displaystyle =\, \dfrac{v\, \pm\, \sqrt{\strut v^2\, -\, 1280\,}}{32}\)

Since this can have only one solution, what then must be true about the expression inside the square root?

Use this fact to find the value of the initial velocity "v", remembering that the cannonball was moving upward, which is a "positive" direction.