Question

A 1kg 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.

Answer 1

Answer:

the position equation of the projectile are

$x = \frac{300}{\sqrt{2} } t +\frac{1}{2} 3t^{2}$

$y= 2+\frac{300}{\sqrt{2} } t-\frac{1}{2} gt^{2}$

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 =$\frac{force}{mass}$

= $\frac{3}{m} =\frac{3}{1}$

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$\frac{m}{s^{2} }$ and in y direction = -g$\frac{m}{s^{2} }$

writing equation of motion in x and y direction:

$x = u_{x} t + \frac{1}{2} at^{2}$

$y = u_{y} t + \frac{1}{2} at^{2}$

$u_{x}$= ucos45 = $\frac{300}{\sqrt{2} }$

$u_{y}$= usin45= $\frac{300}{\sqrt{2} }$

therefore

$x = \frac{300}{\sqrt{2} } t +\frac{1}{2} 3t^{2}$

$y= 2+\frac{300}{\sqrt{2} } t-\frac{1}{2} gt^{2}$

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