How Is the Acceleration of a Rocket Calculated When a Bolt Falls Off?

[Ritzycat]

Homework Statement

A rocket 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 6.0s later.

Homework Equations

$v_x = v_0 + at$
$x_f = x_0 + v_0t + 1/2at^2$
$v_x^2 = v_0^2 + 2a(Δx)$

The Attempt at a Solution

The biggest part I'm confused about is whether or not the bolt had an initial positive velocity (I am setting down as negative). It says the bolt "falls off the side", implying it just falls down.

Position of bolt before it falls
$0m = x + 1/2(-9.8m/s^2)(6.0s)^2$
$x_i = 176.4m$

Acceleration of rocket
$176.4m = 1/2(a)(4.0s)^2$
$a = 22.05 m/s^2$

My answer is incorrect.

 

[Doc Al]

The biggest part I'm confused about is whether or not the bolt had an initial positive velocity (I am setting down as negative). It says the bolt "falls off the side", implying it just falls down.

The bolt will have the same velocity as the rocket at the moment it falls free. "Falls off the side" just means that it's not shot out or launched, but just let go.

 

[PeroK]

or not the bolt had an initial positive velocity (I am setting down as negative). It says the bolt "falls off the side", implying it just falls down.

If you were cycling along and you fell off your bike, would you have forward velocity?

 

[Ritzycat]

Fair enough. I got to go now, but I'll attempt the problem later tonight. Conceptually I understand that - but the wording sort of got to me. Thanks for clarifying!