A ball is projected from top of a tower with a velocity of $5\text{m/s}$ at an angle of ${53}^{\circ }$ to horizontal. Its speed when it is at a height of $0.45\text{m}$ from the point of projection is $(g=10{\text{m/s}}^2)$

The correct option is C $4\text{m/s}$
Given,
projection velocity $u=5\text{m/s}$ and projection angle $\theta ={53}^{\circ }$

The horizontal component of the projection velocity $u_x=5\cos {53}^{\circ }=3\text{m/s}$

And, vertical component of the projection velocity $u_y=5\sin {53}^{\circ }=4\text{m/s}$

At the height $0.45\text{m}$ from the point of projection the vertical component of the velocity is (taking downward negative)
$v_y^2=u_y^2-2gh$
$\Rightarrow v_y^2=(4{)}^2-2\times 10\times 0.45$
$v_y=\sqrt{7}\text{m/s}$

Since, the horizontal component of velocity remain same
$\therefore$ $v_x=u_x=3\text{m/s}$
Hence, $v_{net}=\sqrt{(v_x{)}^2+(v_y{)}^2}=\sqrt{9+7}=4\text{m/s}$

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