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

Answer:

The position vector $\mathbf{r(t) = (260 \sqrt{3}) \overline i + (260 - \dfrac{1}{2} \ gt^2) \ \overline j }$

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

$U_x = 520\ Cos \theta$

$U_x = 520 Cos 30^0$

$U_x = 520 Cos \bigg (\dfrac{\sqrt{3}}{2} \bigg)$

$U_x = 260 Cos \bigg (\sqrt{3} \bigg)$

At the vertical axis

$U_y = 520\ Sin \theta$

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

$U_y = 520\ \times \dfrac{1}{2}$

$U_y =260$

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

$x(t) = U_x(t) + \dfrac{1}{2} at^2$

where; $U_x = 260 \sqrt{3}$   ; a = 0 ; t = t

$x(t) = 260 \sqrt{3}+ 0$

$x(t) = 260 \sqrt{3}$

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

$y(t) = U_y(t) + \dfrac{1}{2}at^2$

where;   $U_y =260$ a = -g

$y(t) =260 t- \dfrac{1}{2}gt^2$

∴

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

$\mathbf{r(t) = (260 \sqrt{3}) \overline i + (260 - \dfrac{1}{2} \ gt^2) \ \overline j }$