An aircraft is traveling horizontally in an easterly direction at a height of 400 m above level ground and at a speed of 120 m/s. when it is directly above the observer, M, it releases a package. In relation to origin O and to x and y directions of east and up respectively, the package has acceleration (-4i -4 j)m/s^2

a) P is the position of the package at time t. Find an expression for the vector MP
b) Find how far from the observer the package lands

u = x (east) velocity = 120 - 4 t

w = z (up)velocity = -4 t

x = 0 + 120 t - 2 t^2
y = 400 + 0 - 2 t^2

when y = 0 the thing lands
t^2 = 200
t = 14.1 s
x = 120(14.1) - 2(14.1)^2
x = 1294

To find the expression for the vector MP, we need to consider the motion of the aircraft and the package separately.

a) The motion of the aircraft can be described by the equation:
x = vt

Where x represents the horizontal distance traveled by the aircraft, v is the velocity of the aircraft, and t is time. Since the aircraft is traveling horizontally in an easterly direction, the velocity vector can be written as v = (v, 0), where v is the speed of the aircraft (120 m/s in this case).

The position of the aircraft when it releases the package is given by the vector:
MA = (x, y)
MA = (v * t, 400)

The acceleration of the package is given as (-4i - 4j) m/s^2. To find the position of the package at time t, we need to consider its motion under this acceleration.

Using the equation of motion, we can find the position of the package as:
r = r0 + v0t + (1/2) a t^2

Given r0 = MA and v0 = 0, the equation becomes:
r = MA + (1/2) a t^2

r = (v * t, 400) + (1/2) (-4i - 4j) t^2

Expanding this expression, we get:
r = (v * t - 2t^2, 400 - 2t^2)

The vector MP is the difference between the position of the package (r) and the position of the observer (M).

MP = r - M
MP = (v * t - 2t^2, 400 - 2t^2) - (0, 0)

Simplifying this, we get:
MP = (v * t - 2t^2, 400 - 2t^2)

Therefore, the expression for the vector MP is MP = (v * t - 2t^2)i + (400 - 2t^2)j.

b) To find how far from the observer the package lands, we need to determine the time at which the package hits the ground. The package will hit the ground when its y-coordinate becomes 0.

So, we can set 400 - 2t^2 = 0 and solve for t:

400 - 2t^2 = 0
2t^2 = 400
t^2 = 200
t = ±√200

Since time cannot be negative, we take the positive square root:
t = √200 = 10√2

Therefore, the package will hit the ground after a time of 10√2 seconds.

To find the horizontal distance from the observer, we can substitute this value of t into the x-coordinate of the vector MP:

x = v * t - 2t^2
x = 120 * (10√2) - 2(10√2)^2
x = 1200√2 - 200 * 2
x = 1200√2 - 400

Hence, the package will land a horizontal distance of 1200√2 - 400 meters from the observer.