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kpg0001 · Jan 25, 2011, 09:15 AM

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!

Unknown008 · Jan 25, 2011, 09:21 AM

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

ebaines · Jan 25, 2011, 12:33 PM

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.

Unknown008 · Jan 25, 2011, 11:00 PM

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