A body is projected at an angle of ${30}^{\circ }$ with the horizontal and with a speed of $30{\text{ms}}^{-1}.$ What is the angle with the horizontal after $1.5\text{seconds}$ ?
$(g=10{\text{ms}}^{-2}).$

The correct option is A $0^{\circ }$
Given, $u=30\text{m/s}$ . Let $\alpha$ be the angle of projectile with the horizontal after $1.5\text{seconds}$.

Let $v$ be the velocity of the projectile after $1.5\text{seconds}$.
$v_x=u_x=u\cos \theta \Rightarrow v_x=30\cos {30}^{\circ }=15\sqrt{3}\text{m/s}$

Using first equation of motion for the vertical component of velocity, we get
$v_y=u_y-gt\Rightarrow v_y=30\sin {30}^{\circ }-10\times 1.5=0$
Vertical component of velocity becomes $0$ when the body is at the highest point of the projectile. $\therefore \alpha =0^{\circ }$

Hence, option $\left(\text{A}\right)$ is the right choice.