Answer:
t = 16.5 s
Explanation:
First we apply first equation of motion to the accelerated motion of the rocket:

where,
vf₁ = final speed of rocket during accelerated motion = ?
vi₁ = initial speed of rocket during accelerated motion = 0 m/s
a = acceleration of rocket during accelerated motion = 30 m/s²
t₁ = time taken during accelerated motion = 4 s
Therefore,

Now, we analyze the motion rocket when engine turns off. So, the rocket is now in free fall motion. Applying 1st equation of motion:

where,
vf₂ = final speed of rocket after engine is off = 0 m/s
vi₂ = initial speed of rocket after engine is off = Vf₁ = 120 m/s
g = acceleration of rocket after engine is off = - 9.8 m/s² (negative sign for upward motion)
t₂ = time taken after engine is off = ?
Therefore,

So, the time taken from the firing position till the stopping position is:

<u>t = 16.5 s</u>
Answer:
it is absorbed or reflected by the atmosphere
Explanation:
In the case when approx 50% only of the solar energy would be directed towards earth and it would be penetrates directly to the surface so the rest or remaining of the radiation would be either absorbed or refected by the atmosphere
So as per the given situation the above represent the answer
hence, the same is to be considered and relevant
A.Momentum Equation
m = mass = 75 kg
v = velocity = 18 m/s
P = momentum
Using the momentum equation , momentum is given as
P = mv
P = 75 x 18
P = 1350 kgm/s
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