Let
denote the rocket's position, velocity, and acceleration vectors at time
.
We're given its initial position

and velocity

Immediately after launch, the rocket is subject to gravity, so its acceleration is

where
.
a. We can obtain the velocity and position vectors by respectively integrating the acceleration and velocity functions. By the fundamental theorem of calculus,


(the integral of 0 is a constant, but it ultimately doesn't matter in this case)

and



b. The rocket stays in the air for as long as it takes until
, where
is the
-component of the position vector.

The range of the rocket is the distance between the rocket's final position and the origin (0, 0, 0):

c. The rocket reaches its maximum height when its vertical velocity (the
-component) is 0, at which point we have


Answer:
733 inches
Step-by-step explanation:
V= 3.14*5^2*28/3
Order of Operations (PEMDAS)
5^2=25
25*3.14= 78.5
78.5*28= 2198
2198/3=732.66
Round to nearest cubic inch
733 inches
Answer:
i would say B/90
Step-by-step explanation:
Hey there! :)
JK ≈ RS
Scale Factor = 9/7
JK = 56
<u>56</u> · <u>9</u> = <u>504</u> = 72
1 7 = 7
Your answer ⇒ B.72
Hope this helps :)
Step-by-step explanation:
(ax + b)² = a²x² + 2abx + b²
In this case, a = 1, so:
14 = 2b
b = 7
(x + 7)² = x² + 14x + 49