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

the position equation of the projectile are

`x = (300)/(√(2) ) t +(1)/(2) 3t^(2)`

`y= 2+(300)/(√(2) ) t-(1)/(2) gt^(2)`

in x and y direction

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 =
`(force)/(mass)`

=
`(3)/(m) =(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
`(m)/(s^(2) )` and in y direction = -g
`(m)/(s^(2) )`

writing equation of motion in x and y direction:

`x = u_(x) t + (1)/(2) at^(2)`

`y = u_(y) t + (1)/(2) at^(2)`

`u_(x)`= ucos45 =
`(300)/(√(2) )`

`u_(y)`= usin45=
`(300)/(√(2) )`

therefore

`x = (300)/(√(2) ) t +(1)/(2) 3t^(2)`

`y= 2+(300)/(√(2) ) t-(1)/(2) gt^(2)`

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