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A particle has initial velocity (2hati+3...

A particle has initial velocity `(2hati+3hatj)ms^(-1)` when it was at origin and has constant acceleration `(3hati+2hatk)ms^(-2)`. Find angle made by displacemnt after 2 sec with XY plane `{sin^(-1)sqrt((k)/(21))}`. Find the value of k.

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The correct Answer is:
2
`vecu=(2hati+3hatj) vecd=vecut+(1)/(2)vecat^(2)`
`veca=(3hati+2hatk)`
Now, if angle is `theta`, then
`theta=sin^(-1)((2)/(21))`
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