Physics Kinematics Problem?

Question

A 1000 kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 14 seconds, then the motor stops. The rocket reaches a maximum altitude 38 seconds after motors exhaust all the fuel. You can ignore any effects of air resistance. (It will help if you draw a vertical motion map and v vs. t graph for this case.) I know that you should break up the problem into two parts (0s to 14s and 14s to 52s), but could use some help putting it together. Thank you for your time.

Answer 1

First of all you don't need the mass, so you don't have to worry about the mass of the rocket changing as the fuel burns (phew! that would require calculus yuck!)

First figure the max velocity (at t=14 sec) by how long it takes to coast to V=0 (at t=38 sec).  Coasting time is 38-14=24 sec, so

Vf = Vo - g * t

0 = Vo - 9.8 * 24; solve for Vo (velocity at t=14 sec):

Vo = 235.2 m/s

The accelleration to get from V=0 (at the ground) to this velocity in 14 sec at constant accelleration is just

V = a * t, solve for a

a = V / t = 235.2 / 14 = 16.8 m/s^2

Note this is the net accelleration; if you needed the thrust you'd have to consider mass and g, but you don't for this problem...)

The max altitude can be found in 2 parts; call the distance covered from t=0-14 sec H1, and the distance covered from t=14-38 sec H2.

For H1:

H1 = Vo * t + 1/2 * a * t^2

H1 = 0 * 14 + 1/2 * 16.8 * 14^2

H1 = 1646.4 m

For H2:

H2 = Vo * t - 1/2 * g * t^2

H2 = 235.2 * 24 - 1/2 * 9.8 * 24^2

H2 = 2822.4 m

Hmax = H1 + H2 = 4468.8 m

Good Luck!