The vertical height in feet of a projectile on a planet in our solar system at a given time t in seconds is represented by the function h(t)=−6t2+24t

Question

. Re-write h(t) in the form \large h\left(t\right)=a\left(t-h\right)^2+k and determine the maximum height of the projectile. Show all work to receive credit for this question.

Answer 1

We want to rewrite the function h(t) in the form h(t) = a(t - h)^2 + k. To do this, we complete the square:

h(t) = -6t^2 + 24t
h(t) = -6(t^2 - 4t)

To complete the square, we need to add and subtract (-4/2)^2 = 4:

h(t) = -6(t^2 - 4t + 4 - 4) (adding and subtracting 4)
h(t) = -6((t - 2)^2 - 4)

Now we can distribute the -6 and simplify:

h(t) = -6(t - 2)^2 + 24

So, a = -6, h = 2, and k = 24. The maximum height of the projectile occurs when t = 2, which is the value of h. Therefore, the maximum height is:

h(2) = -6(2 - 2)^2 + 24 = 24 feet

Answer: \boxed{h(t)=-6(t-2)^2+24}. The maximum height is 24 feet.

Answer 2

in plain mathematical language: .....

h(t)=−6t2+24t
= -6(t^2 - 4t)
= -6(t^2-4t+4 - 4)
= -6( (t - 2)^2 - 4)
= -6(t - 2)^2 + 24

Answer 3

The function h(t) is a quadratic equation in the form ax^2 + bx + c, where a = -6, b = 24, and c = 0. To find the maximum height of the projectile, we need to rewrite h(t) in the form h(t) = a(t - h)^2 + k, where h is the horizontal shift and k is the vertical shift. To do this, we complete the square by adding and subtracting (b/2a)^2, which simplifies to (4/a)^2 = 4. Then, we factor out a from the parentheses and simplify to get the desired form. The maximum height occurs when t = h, so we plug in t = 2, which gives us k = 24, and the maximum height is 24 feet.