Arocket is launched with a constant acceleration straight up. exactly 4.00 seconds after lift off, a bolt falls off the side of the rocket and hits the ground 6.00 seconds later. what was the rockets acceleration?

The initial speed of the bolt is not 58.86 m/s.  

Let a be the acceleration of the rocket.  

During the 4 sec lift off, the rocket has reached a height of  

h = (1/2)*a*t^2  

with t=4,  

h = (1/2)*a^16  

h = 8*a  

Its velocity at 4 sec is  

v = t*a  

v = 4*a  

The initial velocity of the bolt is thus 4*a.  

During the 6 sec fall, the bolt has the initial velocity V0=-4*a and it drops a total height of h=8*a. From the equation of motion,  

h = (1/2)*g*t^2 + V0*t  

Substituting h0=8*a, t=6 and V0=-4*a into it,  

8*a = (1/2)*g*36 - 4*a*6  

Solving for a  

a = 5.52 m/s^2

i think the answer would be b