A ball is projected from top of a tower with a velocity of $5m/s$ at an angle of ${53}^o$ to horizontal. Its speed when it is at a height of $0.45m$ from the point of projection is:

The correct option is C $4m/s$
Initial vertical speed is $u=5m/s\times sin{53}^0=4m/s$

so if the final $vertical$ velocity being $v$ at a height $h$ then $v^2=u^2-2gh$ will give us $v=\sqrt{16-20\times 0.45}=\sqrt{7}m/s$

the $horizontal$ velocity remain constant and given as $V_x=5m/sCos{53}^0=3m/s$

so the speed at the said instant will be $V=\sqrt{V_x^2+v^2}=\sqrt{9+7}=4m/s$