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.

A particle has initial velocity (2hati+3hatj)ms^(-1) when it was at or

1.	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.
Answer» Solution :`vecu=(2hati+3hatj) vecd=vecut+(1)/(2)vecat^(2)` `VECA=(3hati+2hatk)` Now, if ANGLE is `THETA`, then `theta=sin^(-1)((2)/(21))`