I should be able to answer this having taken the class, but I'm drawing a blank.

A rocket takes off with constant acceleration.
After 4 seconds a bolt falls off the rocket.
The bolt hits the ground 6.3 seconds after falling off.
What is the acceleration of the rocket?

Help would be much appreciated!

Use the kinematics formulae.

The height h above the ground of the rocket is given by:

h = \frac12 at^2

coming from the basic formula s = s_o + ut + \frac12 at^2 but then, both so and u = 0.

So, after 4 seconds, the rocket is at height h1.

The time the bolt takes to reach the ground is 6.3 seconds and is given by:

h = \frac12 gt^2

You know g, you know t (6.3), find the value of h.

Use this h to get the value of a in the first formula, knowing t (4).

EDIT: Forgot the velocity of the rocket :o

h =ut + \frac12 gt^2

where u = at

Not quite. The equation of motion for the bolt is:


s = s_0 + ut + \frac 1 2 a t^2


Where s_0 is the height of the rocket and u is the velocity of the racket at t=4 seconds, and a is -g. So:


s_0 = \frac 1 2 a t^2


where a = acceleration of rocket and t = 4 seconds, and


u = at


Hence for the bolt:


0 = \frac 1 2 a (4s)^2 + a(4s)(6s) - \frac 1 2 g (6s)^2

Solve for a.

Oh, right I forgot about the initial vertical speed of the rocket >.<