A ball is acted upon by the following velocities. (i) `3 ms^(-1)` due East 9ii) `11 ms^(-1)` due South and (iii) `5 sqrt 2 ms^(-1)` due North-East. Find the magnitude and direction recultant velocity.

Explanation

The v arious velocities acting on the ball are shown in Fig. 2 (c ) .59 (a0. `vec (OP) =3 ms^(-1)`
due East, `vec (OQ) =1 ms^(-10` due South an `
`vec (OS) =5 sqrt 2 ms^(-1)` inclined at an angle `45^(0)` with East.
Resolving `vec (OS)` into two rectangular components along East and North directions, we get
.
`5 sqrt 2 cos 45 ^(0) =5 ms^(-1) ` along East
and `5 sqrt 2 sin 45^(0) =5 ms^(-1)` along Northe
:. Effective velocity of ball along East,
`A= 3+ 5 =8 m//s`, represented by `vec (OP)'`
and the effective velcoity of ball along South,
`B=11 -5 =6 m//s`, represented by`vec (OQ')`.
The resultant velcity `vec (R)` will be represented by `vec (OT)`, where `/_P'OQ' =90^(0)`. Fig. 2 (c ) .59 (b).
Thus magnitude of `vec (R)` will be `
`R= (sqrt A^(2) +B^(2) ) =sqrt(8^(2) +6^(2)) + =10 ms^(-1)`
Let`vec R` make and angle `beta` with East, then `
`tan `beta =(B)/(A) =6/8 =0.75 =tan =tan 36 &(0) 52'`
or `beta =36^(0) 52' South of East `.