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
20m away
the dog was 60m away from. you subtract 40m since it is 10m/s x 4 seconds
