Answer:
Initial horizontal velocity: approximately [tex]9.8\; {\rm m\cdot s^{-1}}[/tex].
Assuming that air resistance is negligible and [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex], horizontal acceleration would be [tex]0\; {\rm m\cdot s^{-2}}[/tex] and vertical acceleration would be: [tex](-9.81)\; {\rm m\cdot s^{-2}}[/tex].
Initial vertical velocity: approximately [tex](-6.9)\; {\rm m\cdot s^{-1}}[/tex] (downward.)
Explanation:
Let [tex]u = 12\; {\rm m\cdot s^{-1}}[/tex] denote the magnitude of the initial velocity. Let [tex]\theta = 35^{\circ}[/tex] denote the angle between this initial velocity vector and the horizontal direction.
Refer to the diagram attached (not drawn to scale.) Initial velocity [tex]u[/tex], its horizontal component [tex]u_{x}[/tex], and its vertical component [tex]u_{y}[/tex] form a right triangle:
Hence, the horizontal and vertical component of initial velocity would be:
[tex]\begin{aligned}u_{x} &= (\text{adjacent}) \\ &= (\text{hypotenuse})\, \frac{(\text{adjacent})}{(\text{hypotenuse})} \\ &= u\, \cos(\theta) \\ &= (12\, \; {\rm m\cdot s^{-1}})\, \cos(35^{\circ}) \\ &\approx 9.8\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
[tex]\begin{aligned}u_{y} &= (\text{opposite}) \\ &= (\text{hypotenuse})\, \frac{(\text{opposite})}{(\text{hypotenuse})} \\ &= u\, \sin(\theta) \\ &= (12\, \; {\rm m\cdot s^{-1}})\, \sin(35^{\circ}) \\ &\approx 6.9\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
Optionally, add a negative sign to [tex]u_{y}[/tex] to show that the vertical component of initial velocity points downwards: [tex](-6.9)\; {\rm m\cdot s^{-1}}[/tex].
If air resistance on the book is negligible, the only force on the book during the flight would be gravitational attraction from the planet. The acceleration of the book would be in the same direction as gravitational attraction, which points downwards in the vertical direction:
