Question

A battleship fires a heavy caliber gun towards a fighter aircraft that flies at an unknown altitude. The maximum speed of a bullet is 520 m/s and height to the exit of the riffle is 9 m. The gun is kept at angle of elevation of 30o .( Ignore air resistance.) (a) Write the position vector r(t)

Answer 1

The position vector
`\mathbf{r(t) = (260 √(3)) \overline i + (260 - (1)/(2) \ gt^2) \ \overline j }`

Step-by-step explanation:

Given that:

The maximum speed of the projection of the bullet = 520 m/s

The height to the exit of the rifle = 9 m

The angle of the elevation = 30°

Taking the vector position with respect to the horizontal axis and vertical axis; we have:

Initial Velocity:

At the horizontal axis

`U_x = 520\ Cos \theta`

`U_x = 520 Cos 30^0`

`U_x = 520 Cos \bigg ((√(3))/(2) \bigg)`

`U_x = 260 Cos \bigg (√(3) \bigg)`

At the vertical axis

`U_y = 520\ Sin \theta`

`U_y = 520\ Sin \bigg (30^0 \bigg)`

`U_y = 520\ * (1)/(2)`

`U_y =260`

The displacement in the (x) direction can be computed as:

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

where;
`U_x = 260 √(3)` ; a = 0 ; t = t

`x(t) = 260 √(3)+ 0`

`x(t) = 260 √(3)`

The displacement in the (y) direction can be computed as:

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

where;
`U_y =260` a = -g

`y(t) =260 t- (1)/(2)gt^2`

∴

The position vector r(t) =
`x \overline i + y \overline j`

`\mathbf{r(t) = (260 √(3)) \overline i + (260 - (1)/(2) \ gt^2) \ \overline j }`