A boy is running along positive x - axis with `9 m//s`. While running he manages to throw a stone in a plane perpendicular to his direction of running with velocity `12 m//s` at an angle `30^@` with vertical. Find the speed of the stone at the highest point of the trajectory.

Explanation

Initial velocity of stone
`vec u _("stone") = 8 hat i + (12 cos 30 hat j) + (12 sin 30 hat k)(m//s)`
=`8 hat i + 6 sqrt(3) hat j + 6 hat k (m//s)`
At the highest point, vertical component of velocity will become zero.
So, `vec v_(highest) = 8 hat i+ 6 hat k (m//s)`
Hence magnitude of the velocity at highest point
`|vec v_(highest)| = sqrt(8^2 + 6^2)) = 10 m//s`.
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