Answer
$6.3 m/s,\space \theta=18.43^{\circ}$ North of east
Work Step by Step
Please see the attached image first.
Let's take,
Velocity of the hawk relative to the balloon = $V_{HB}$
Velocity of the balloon relative to the ground = $V_{BG}$
Velocity of the hawk relative to the ground = $V_{HG}$
By using the Pythagorean theorem, we can get.
$V_{HG}=\sqrt {V_{HB}^{2}+V_{BG}^{2}}$ ; Let's plug known values into this equation.
$V_{HG}=\sqrt{(2\space m/s)^{2}+(6\space m/s)^{2}}=6.3\space m/s$
By using trigonometry, we can get.
$tan\theta=\frac{V_{HB}}{V_{BG}}=\frac{2\space m/s}{6\space m/s}=>\theta=tan^{-1}(0.33)=18.43^{\circ}$ North of east
