Answer:
A. Vx = 3.63 m/s
B. Vy = -45.73 m/s
C. |V| = 45.87 m/s
D. θ = -85.46°
Explanation:
Given that position, r, is given as:
r = 3.63tˆi − 5.73t^2ˆj + 8.16ˆk
Velocity is the derivative of position, r:
V = dr/dt = 3.63 - 11.46t^j
A. x component of velocity, Vx = 3.63 m/s
B. y component of velocity, Vy = -11.46t
t = 3.99 secs,
Vy = - 11.46 * 3.99 = -45.73 m/s
C. Magnitude of velocity, |V| = √[(-45.73)² + 3.63²]
|V| = √(2091.2329 + 13.1769)
|V| = √(2104.4098)
|V| = 45.87 m/s
D. Angle of the velocity relative to the x axis, θ is given as:
tanθ = Vy/Vx
tanθ = -45.73/3.63
tanθ = -12.6
θ = -85.46°
1) The distance travelled by the rocket can be found by using the basic relationship between speed (v), time (t) and distance (S):

Rearranging the equation, we can write

In this problem, v=14000 m/s and t=150 s, so the distance travelled by the rocket is

2) We can solve the second part of the problem by using the same formula we used previously. This time, t=300 s, so we have:

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26) A) 400J as GPE = mgh
27) C) Both obj has same PE
28) B) It decreases as energy is not being provided.
29) C) (42 - 0)/7 = 6m/s2