Respuesta :
Answer:
the position equation of the projectile are
[tex]x = \frac{300}{\sqrt{2} } t +\frac{1}{2} 3t^{2}[/tex]
[tex]y= 2+\frac{300}{\sqrt{2} } t-\frac{1}{2} gt^{2}[/tex]
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 =[tex]\frac{force}{mass}[/tex]
= [tex]\frac{3}{m} =\frac{3}{1}[/tex]
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[tex]\frac{m}{s^{2} }[/tex] and in y direction = -g[tex]\frac{m}{s^{2} }[/tex]
writing equation of motion in x and y direction:
[tex]x = u_{x} t + \frac{1}{2} at^{2}[/tex]
[tex]y = u_{y} t + \frac{1}{2} at^{2}[/tex]
[tex]u_{x}[/tex]= ucos45 = [tex]\frac{300}{\sqrt{2} }[/tex]
[tex]u_{y}[/tex]= usin45= [tex]\frac{300}{\sqrt{2} }[/tex]
therefore
[tex]x = \frac{300}{\sqrt{2} } t +\frac{1}{2} 3t^{2}[/tex]
[tex]y= 2+\frac{300}{\sqrt{2} } t-\frac{1}{2} gt^{2}[/tex]
here 2 is added as the projectile already 2 meter above the ground
