A bullet with muzzle velocity $100\sqrt{2}\text{m/s}$ is to be fired at an angle of ${45}^{\circ }$ with the horizontal. Find the coordinates of bullet after $t=4\text{s}$, assuming bullet was initially at origin. Take $g=10{\text{m/s}}^2$.

The correct option is A $400\text{m},320\text{m}$
Given;
$u=100\sqrt{2}\text{m/s}$
$\theta ={45}^{\circ }$ from horizontal
$t=4\text{s}$
Reference : Upward vertical direction is taken $+ve$
$a_y=-g$
For horizontal direction:
$X=u_x\times t=u\cos \theta \times t$
$X=100\sqrt{2}\times \cos {45}^{\circ }\times 4$
$\therefore X=400\text{m}$

For vertical direction:
$Y=u_y\times t-\frac{1}{2}{a}_y{t}^2$
$Y=u\sin {45}^{\circ }\times 4-\frac{1}{2}\times 10\times 4^2$
$=100\sqrt{2}\times \frac{1}{\sqrt{2}}\times 4-80$
$=400-80$
$\therefore Y=320\text{m}$
Coordinates of bullet at $t=4\text{s}$
$(X,Y)=(400\text{m},320\text{m})$