Arocket is launched straight up with constant acceleration. four seconds after liftoff, a bolt falls off the side of the rocket. the bolt hits the ground 6.0 s later. what was the rocket’s acceleration?

5.5 m/s²

Explanation:

The rocket has initial velocity 0 m/s, and acceleration a.  After 4 seconds, it has a velocity of v and position h.

When the bolt falls off, it has initial velocity v and acceleration -9.8 m/s².  After 6 seconds, it falls a distance of h.

For the rocket:

Δy = ½ (v + v₀) t

h = ½ (v + 0) (4)

h = 2v

For the bolt:

Δy = v₀ t + ½ at²

-h = v (6) + ½ (-9.8) (6)²

h = 176.4 − 6v

Substitute:

2v = 176.4 − 6v

8v = 176.4

v = 22.05

Therefore, the rocket's acceleration is:

a = (v − v₀) / t

a = (22.05 − 0) / 4

a = 5.5 m/s²

c. 7/3

step-by-step explanation:

a rational number is like pi- it is a repeating decimal.

a. is -1.8

b. 2.236068

c. 2.3333333333333333333

d.   is 3

i hope that ! =)