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
P = m * v
P = 100 * 5 Kg-m/s towards south
P = 500 Kg-m/s towards south
In short, your answer would be 500 Kg-m/s towards south
Hope this helps!
D.
The molecule escapes the ocean via evaporation and then becomes part of a cloud via condensation.
