PLEASE HELP ME IM ON A TIMER
1 answer:
Answer:
Displacement: 6.71 m, Direction: 63.4 degrees north of east
Explanation:
In the attached image we can aprecciate each one of the movements of the parade. Let's say that the parade started from the origin (point (0,0)) then it moves to the east 4 blocks it means now the parade is located at point (4,0).
Then the parade went to the south three blocks, so it moves to the coordinate (4,-3). After this the parade went to the west one block so the new coordinate point is (3, -3).
And finally the movement of the 0 parade was 9 blocks to the north. It means the final point is now (0,9) - (3,-3) = (3,6)
And the displacement will be defined by the folliwing vector operation:
We know that the magnitude of the displacement vector is defined by the phytagoras theorem
And the angle will be defined by:
tan(beta)=3/6
beta = tan^-1(6/3)
beta = 63.43°
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Answer:
(a) r = 1.062·R =
(b) r =
(c) Zero
Explanation:
Here we have escape velocity v given by
and the maximum height given by
Therefore, when the initial speed is 0.241v we have
v = so that;
v² =
v² =
is then
Which gives
or
r = 1.062·R
(b) Here we have
Therefore we put in the maximum height equation to get
From which we get
r = 1.32·R
(c) The we have the least initial mechanical energy, ME given by
ME = KE - PE
Where the KE = PE required to leave the earth we have
ME = KE - KE = 0
The least initial mechanical energy to leave the earth is zero.
1) Distance down the hill: 1752 ft (534 m)
2) Time of flight of the shell: 12.9 s
3) Final speed: 326.8 ft/s (99.6 m/s)
Explanation:
1)
The motion of the shell is a projectile motion, so we can analyze separately its vertical motion and its horizontal motion.
The vertical motion of the shell is a uniformly accelerated motion, so the vertical position is given by the following equation:
(1)
where:
is the initial vertical velocity of the shell, with
and
is the acceleration of gravity
At the same time, the horizontal motion of the shell is a uniform motion, so the horizontal position of the shell at time t is given by the equation
where is the initial horizontal velocity of the shell.
We can re-write this last equation as
(1b)
And substituting into (1),
(2)
where we have choosen the top of the hill (starting position of the shell) as origin (0,0).
We also know that the hill goes down with a slope of from the horizontal, so we can write the position (x,y) of the hill as
(3)
Therefore, the shell hits the slope of the hill when they have same x and y coordinates, so when (2)=(3):
Substituting (1b) into this equation,
Which has 2 solutions:
x = 0 (origin)
and
So, the distance d down the hill at which the shell strikes the hill is
2)
In order to find how long the mortar shell remain in the air, we can use the equation:
where:
x = 1322 ft is the final position of the shell when it strikes the hill
is the initial velocity of the shell
is the angle of projection of the shell
Substituting these values into the equation, we find the time of flight of the shell:
3)
In order to find the final speed of the shell, we have to compute its horizontal and vertical velocity first.
The horizontal component of the velocity is constant and it is
Instead, the vertical component of the velocity is given by
And substituting at t = 12.9 s (time at which the shell strikes the hill),
Therefore, the final speed of the shell is:
Learn more about projectile motion:
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