Answer
(a) The speed relative to the ground is 8.11 m/s.
(b) The magnitude of the acceleration is $9.48~m/s^2$ and it is directed horizontally toward the center of the circle of rotation.
(c) The angle between the bird's velocity vector and the horizontal is $21.7^{\circ}$.
Work Step by Step
(a) $v_x = \frac{2\pi ~r}{t} = \frac{(2\pi)(6.00~m)}{5.00~s}$
$v_x = 7.54~m/s$
$v_y = 3.00~m/s$
We can find the speed relative to the ground.
$v = \sqrt{(7.54~m/s)^2+(3.00~m/s)^2}$
$v = 8.11~m/s$
The speed relative to the ground is 8.11 m/s.
(b) $a_x = \frac{v_x^2}{r} = \frac{(7.54~m/s)^2}{6.00~m}$
$a_x = 9.48~m/s^2$
The upward velocity is constant, so $a_y = 0$
The magnitude of the acceleration is $9.48~m/s^2$ and it is directed horizontally toward the center of the circle of rotation.
(c) $tan(\theta) = \frac{3.00~m/s}{7.54~m/s}$
$\theta = tan^{-1}(\frac{3.00}{7.54}) = 21.7^{\circ}$.
The angle between the bird's velocity vector and the horizontal is $21.7^{\circ}$.
