A ball is projected with a velocity, 10 ms^–1, at an angle of 60° with the vertical direction. Its speed at the highest point of its trajectory will be:

$$\begin{gathered} \mathrm{Angle} \mathrm{of} \mathrm{projection} \mathrm{with} \mathrm{vertical} \mathrm{is} {60}^0 \\ \mathrm{so} \mathrm{with} \mathrm{horizontal} \mathrm{it} \mathrm{is} {30}^0 \\ \mathrm{velocity} \mathrm{at} \mathrm{the} \mathrm{highest} \mathrm{point} \mathrm{will} \mathrm{be} \mathrm{its} \\ \mathrm{horizontal} \mathrm{velocity} \mathrm{ucos}{30}^0 \end{gathered}$$

$10\cos {30}^0=5\sqrt{3 } \mathrm{m}/\mathrm{s}$