A1kg projectile is launched from a platform 2m above ground northwards with initial speed of 300m/s and an angle of elevation of π 4 above the horizon. if the wind applies a force of 3n to the east, find the position function of the object.

the position equation of the projectile are

in x and y direction

Step-by-step explanation:

let the mass of the projectile be m, initial velocity be u

the wind applies a force of 3 newton in east direction.

therefore acceleration due to the force in east direction =

=

acceleration due to gravity is in south direction = g

let east be x direction and north be y direction.

therefore acceleration in x direction = 3 and in y direction = -g

writing equation of motion in x and y direction:

= ucos45 =

= usin45=

therefore

here 2 is added as the projectile already 2 meter above the ground

1)x=-6,y=1 2)a=0,b=-8 3)j=8,k=1 4)m=3,n=-6

Step-by-step explanation:

1)x-4y=-10,2x+y=-11

x-4y=-10

x=4y-10

2\left(4y-10\right)+y=-11

8y-20+y=-11

9y-20=-11

9y=9

y=1

x=4-10

x=-6

x=-6,y=1

2)a-3b=24,a+2b=-16

a-3b=24

a=3b+24

3b+24+2b=-16

5b+24=-16

5b=-40

b=-8

a=3\left(-8\right)+24

a=-24+24

a=0

a=0,b=-8

3)k-j=-7

4j-5k=27,-j+k=-7

4j-5k=27

4j=5k+27

j=\frac{1}{4}\left(5k+27\right)

j=\frac{5}{4}k+\frac{27}{4}

-\left(\frac{5}{4}k+\frac{27}{4}\right)+k=-7

-\frac{1}{4}k-\frac{27}{4}=-7

-\frac{1}{4}k=-\frac{1}{4}

k=1

j=\frac{5+27}{4}

j=8

j=8,k=1

4)m+n=-3,m-4n=27

m+n=-3

m=-n-3

-n-3-4n=27

-5n=30

n=-6

m=-\left(-6\right)-3

m=6-3

m=3

m=3,n=-6