Thank you for reporting, we will resolve it shortly
Q.
A ball is projected with a velocity, $10\, ms ^{-1}$, at an angle of $60^{\circ}$ with the vertical direction. Its speed at the highest point of its trajectory will be :
At highest point only horizontal component of velocity remains
$\Rightarrow u_{x}=u \cos \theta$
$ u _{ x }= u \cos \theta =10 \cos 30^{\circ} $
$=5 \sqrt{3} \,ms ^{-1}$