Homework Problem Please Help, projectile motion

1. The problem statement, all variables and given/known data
A model rocket blasts off from the ground, rising straight upward with a constant acceleration that has a magnitude of 79.0 m/s2 for 1.90 seconds, at which point its fuel abruptly runs out. Air resistance has no effect on its flight. What maximum altitude (above the ground) will the rocket reach?

2. S=.5at2^2

3. The attempt at a solution

S= .5(79)(1.90)2
=142.595

S=142.595^2/(2*9.8)
=1037.42

142.595+1037.42=1080m.... only 3 sig figs

It says its wrong?

For the second stage when the rocket is decelerating you need to work out the initial speed, which is the speed just when the fuel runs out. You know the final speed (0) and you know the acceleration (-g) and once you've worked the initial speed you can use v^2 = u^2 + 2 a s to work out the distance.

For this problem I am going to use the two equations which involve constant acceleration.
s_final = (1/2)a t^2 + v_0 t + s_0 AND
v_final = a t + v_0
That's 2 equations, with six unknowns (a, t, s_0, v_0, s_final, v_final) so to complete solve for the motion, we need four pieces of information.

Here we break up the problem into two phases of constant acceleration -- the initial acceleration up and the free-fall segment, where only gravity operates.

Here's a hint to prevent you from making mistakes when working with complicated problems. Try keeping your units with your numbers.

Phase 1:
s_0 = 0 (Start on the ground)
v_0 = 0 (Start at rest)
t = 1.90s (Given)
a = 79.0 m/s^2 (Given)

v_final = a t + v_0 = (79.0 m/s^2)(1.90 s)
s_final = a t^2/2 + v_0 t + s_0 = (0.5)(79.0 m/s^2)(1.90 s)(1.90 s)

This part you seem to have no trouble with.

Phase 2: The rocket keeps on going, but now the constant acceleration is gravity

s_0 = (0.5)(79.0 m/s^2)(1.90 s)(1.90 s) (copied from Phase 1)
v_0 = (79.0 m/s^2)(1.90 s) (copied from Phase 1)
a = -9.81 m/s^2 (Gravity -- although if you are told to use a different number than you should use it)
v_final = 0 (This is the condition of being at the top of a free-fall motion)

v_final = a t + v_0 =>t = (v_final - v)/(a) = -(79.0 m/s^2)(1.90 s)/(-9.81 m/s^2)
s_final = a t^2/2 + v_0 t + s_0 = (0.5)(-9.81 m/s^2) (-(79.0 m/s^2)(1.90 s)/(-9.81 m/s^2))^2 + (79.0 m/s^2)(1.90 s)(-(79.0 m/s^2)(1.90 s)/(-9.81 m/s^2)) + (0.5)(79.0 m/s^2)(1.90 s)(1.90 s)


Since I posted, I read that prometheus helps with v^2 = u^2 + 2 a s which means in my notation:
(v_final)^2 = (v_0)^2 + 2 a (s_final - s_0)

And you can get this with algebra from
s_final = (1/2)a t^2 + v_0 t + s_0 AND
v_final = a t + v_0 => t = (v_final - v_0)/a
=> s_final - s_0 = (1/2)a [ (v_final - v_0)/a]^2 + v_0 [ (v_final - v_0)/a ]
=> 2(s_final - s_0) = ((v_final)^2 - 2(v_final)(v_0) + (v_0)^2)/a + (2(v_final)(v_0) - (v_0)^2)/a
=> 2a(s_final - s_0) = (v_final)^2 - (v_0)^2
=> (v_final)^2 = (v_0)^2 + 2a(s_final - s_0) QED

Using it for phase 2 gives the same answer, but without any need to write out the expression for t:
(v_final)^2 = (v_0)^2 + 2a(s_final - s_0)
0^2 = [(79.0 m/s^2)(1.90 s)]^2 + 2(-9.81 m/s^2)(s_final - [(0.5)(79.0 m/s^2)(1.90 s)(1.90 s)])
=> s_final = (0.5)(79.0 m/s^2)(1.90 s)(1.90 s) + -[(79.0 m/s^2)(1.90 s)]^2 / ( 2(-9.81 m/s^2) ) = 142.595 m + 1148.319 m = 1290.914 m

This post has been edited by rpenner on Sep 20 2009, 07:58 PM