Answers:
a) -171.402 m/s
b) 17.49 s
c) 1700.99 m
Explanation:
We can solve this problem with the following equations:
(1)
(2)
(3)
Where:
is the bomb's final height
is the bomb's initial height
is the bomb's initial vertical velocity, since the airplane was moving horizontally
is the time
is the acceleration due gravity
is the bomb's range
is the bomb's initial horizontal velocity
is the bomb's final velocity
Knowing this, let's begin with the answers:
<h3>b) Time
</h3>
With the conditions given above, equation (1) is now written as:
(4)
Isolating
:
(5)
(6)
(7)
<h3>a) Final velocity
</h3>
Since
, equation (3) is written as:
(8)
(9)
(10) The negative sign only indicates the direction is downwards
<h3>c) Range
</h3>
Substituting (7) in (2):
(11)
(12)
Answer:
Speed at which the ball passes the window’s top = 10.89 m/s
Explanation:
Height of window = 3.3 m
Time took to cover window = 0.27 s
Initial velocity, u = 0m/s
We have equation of motion s = ut + 0.5at²
For the top of window (position A)

For the bottom of window (position B)


We also have

Solving

So after 1.11 seconds ball reaches at top of window,
We have equation of motion v = u + at

Speed at which the ball passes the window’s top = 10.89 m/s
I am pretty sure that <span>the following whihc cannot be determined by looking at the phase diagram is definitely </span>D. system pressure. I consider this one to be correct because only this point is not included into<span> phase diagram and can't be determined itself. Hope it will help! Regards!</span>
Increasing mass increases kinetic energy. This can be seen in the equation KE = 1/2 (m) (v)^2
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