Explanation:
Let i, j and k represents east, north and upward direction respectively.
Velocity due north, 
Velocity of the crosswind, 
Velocity of downdraft, 
(downward direction)
(a) Let v is the position vector that represents the velocity of the plane relative to the ground. It is given by :
(b) The speed of the plane relative to the ground can be calculated as :
v = 66 m/s
Hence, this is the required solution.
Answer:
jumping from cliff sounds nice
Explanation:
Answer:
t = 13.43 s
Explanation:
In order to find the minimum time required by the plane to stop, we will use the first equation of motion. The first equation of motion is written as follows:
Vf = Vi + at
where,
Vf = Final Velocity of the Plane = 0 m/s (Since, the plane finally stops)
Vi = Initial Velocity of the Plane = 95 m/s
a = deceleration of the plane = - 7.07 m/s²
t = minimum time interval needed to stop the plane = ?
Therefore,
0 m/s = 95 m/s + (- 7.07 m/s²)t
t = (95 m/s)/(7.07 m/s²)
<u>t = 13.43 s</u>
Yes, but it's a null vector .So,no vector can have a component at right angle to itself unless it is a zero vector.
a₀). You know ...
-- the object is dropped from 5 meters
above the pavement;
-- it falls for 0.83 second.
a₁). Without being told, you assume ...
-- there is no air anyplace where the marshmallow travels,
so it free-falls, with no air resistance;
-- the event is happening on Earth,
where the acceleration of gravity is 9.81 m/s² .
b). You need to find how much LESS than 5 meters
the marshmallow falls in 0.83 second.
c). You can use whatever equations you like.
I'm going to use the equation for the distance an object falls in
' T ' seconds, in a place where the acceleration of gravity is ' G '.
d). To see how this all goes together for the solution, keep reading:
The distance that an object falls in ' T ' seconds
when it's dropped from rest is
(1/2 G) x (T²) .
On Earth, ' G ' is roughly 9.81 m/s², so in 0.83 seconds,
such an object would fall
(9.81 / 2) x (0.83)² = 3.38 meters .
It dropped from 5 meters above the pavement, but it
only fell 3.38 meters before something stopped it.
So it must have hit something that was
(5.00 - 3.38) = 1.62 meters
above the pavement. That's where the head of the unsuspecting
person was as he innocently walked by and got clobbered.
