(a) 3675 N
Assuming that the acceleration of the rocket is in the horizontal direction, we can use Newton's second law to solve this part:
where
is the horizontal component of the force
m is the mass of the passenger
is the horizontal component of the acceleration
Here we have
m = 75.0 kg
Substituting,
(b) 3748 N, 11.3 degrees above horizontal
In this part, we also have to take into account the forces acting along the vertical direction. In fact, the seat exerts a reaction force (R) which is equal in magnitude and opposite in direction to the weight of the passenger:
where we used
as acceleration of gravity.
So, this is the vertical component of the force exerted by the seat on the passenger:
and therefore the magnitude of the net force is
And the direction is given by