A model rocket is launched straight upward from the side of a 256-ft cliff. The initial velocity is 96 ft/sec. The height of the rocket h(t) is given by:
h(t) =-16t^2+96t+256
where h(t) is measured in feet and t is the time in seconds. Determine the time at which the rocket is at the maximum height and the maximum height it reaches.
To determine the time at which the rocket is at the maximum height and the maximum height it reaches, we can use the formula for finding the vertex of a quadratic function. In this case, the function h(t) represents the height of the rocket as a function of time.
The vertex of a quadratic function in the form f(x) = ax^2 + bx + c is given by the coordinates (h, k), where:
h = -b / (2a)
k = f(h)
In our case, a = -16, b = 96, and c = 256. Let's calculate the values of h and k to find the time and maximum height of the rocket.
First, we calculate h using the formula:
h = -b / (2a)
= -96 / (2(-16))
= -96 / (-32)
= 3
Next, we calculate k by substituting h into the equation:
k = f(h)
= -16h^2 + 96h + 256
= -16(3)^2 + 96(3) + 256
= -16(9) + 288 + 256
= -144 + 288 + 256
= 400
Therefore, the rocket reaches its maximum height at time t = 3 seconds, and the maximum height it reaches is 400 feet.
to find the vertex, do -b/2a which gives you 3 (your x value of the vertex). Plug in 3 to get the y value.
x=seconds
y=time
Try factoring the quadratic with 128 and 32 to find what "t" is
that is a parabola, where is the vertex?
16 t^2 - 96 t = -h +256
t^2 - 6 t = (1/16)(-h + 256)
t^2 - 6 t + (6/2)^2 = (1/16)(-h + 256) + 9
(t-3)^2 = (1/16)(-h + 256) +(1/16)(144)
(t-3)^2 = -(1/16)(h-400)
vertex at 3 seconds and h =400 ft
h(t) =-16t^2+96t+256
= -16(t^2-6t-16)
so the roots (h=0) are
(t-8)(t+2)=0, or t=-2, and 8
Becuse this is a parabola, the maximum will occur when t is halfway between, or t=3 (check that).
so the max height occurs when t = 3, solve for h(3).