(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:
[tex]F_x = m a_x[/tex]
where
[tex]F_x[/tex] is the horizontal component of the force
m is the mass of the passenger
[tex]a_x[/tex] is the horizontal component of the acceleration
Here we have
m = 75.0 kg
[tex]a_x = 49.0 m/s^2[/tex]
Substituting,
[tex]F_x=(75.0)(49.0)=3675 N[/tex]
(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:
[tex]R=mg=(75.0)(9.8)=735 N[/tex]
where we used
[tex]g=9.8 m/s^2[/tex] as acceleration of gravity.
So, this is the vertical component of the force exerted by the seat on the passenger:
[tex]F_y = 735 N[/tex]
and therefore the magnitude of the net force is
[tex]F=\sqrt{F_x^2+F_y^2}=\sqrt{3675^2+735^2}=3748 N[/tex]
And the direction is given by
[tex]\theta = tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{735}{3675})=11.3^{\circ}[/tex]
