Afree-falling projectile is launched from the ground and has a hangtime of 6 seconds before hitting the ground. how high did it go?

1A)  

x=v0x*t=v0cosθ*t  

x=52co31*3.2=142.6 m  

1B)  

y0=1/2gt^2-v0y*t=1/2gt^2-v0sinθt  

y=0.5*9.8*3.2^2-52*sin31*3,2=23.4 m  

2A)  

x=2v0^2sin(2θ)/g  

v0=[xg/2sin(2θ)]^1/2=14.4 m/s  

the initial speed relative to the ground is  

v=v0-4.4=10 m/s  

2B)  

fly time is  

t=2voy/g  

t=2*14.4/9.8=2.94  

2C)  

mgy=1/2mv0y^2  

y=v0y^2/(2g)=10.58 m

A.) 3605.6 N

B.) 33.7 degree

Explanation:

To find the result force acting on the wing of the airplane, we need to resolve the forces into x and y components

Resolving into x component :

Sum of forces = 3500 - 500 = 3000N

Resolving into y component:

Sum of forces = 2000N

Resultant force Fr = sqrt ( Fx^2 + Fy^2)

Fr = sqrt ( 3000^2 + 2000^2 )

Fr = sqrt ( 9000000 + 4000000 )

Fr = sqrt ( 13000000)

Fr = 3605.6 N

Therefore, resultant force acting on the wing is 3605.6 N

The direction of the vector will be:

Tan Ø = Fy / Fx

Substitute Fx and Fy into the formula

Tan Ø = 2000 / 3000

Tan Ø = 0.66666

Ø = tan^-1(0. 66666)

Ø = 33.7 degree.