Answer this...?

Question

A pollution-sampling rocket is launched straight upward with rockets providing a constant acceleration of 12.0m/s^2 for the 1st 1000m of flight. At that pt the rocket itself is in free fall. Whats the rockets speed when the engine cut off? Whats the max altitude reached by the rocket? Whats the time taken to get the max altitude?

Answer 1

x = 0.5*at^2
t(1000) = (2x/a)^0.5 = 12.9 sec
v(1000) = at = 155 m/sec

v(1000) - vf = -gt
t(max) = 155/9.8 = 15.8 sec
x(max) = 1000+ v(1000)t + 0.5*gt^2 = 2230 m

The t(max) above is the time from engine cutoff to max altitude. The total time is 15.8 + 12.9 = 28.7 sec.

Answer 2

Do your own homework!

Answer 3

a = 12m/s^2
h1 = 1000m
-----------------
v = sqrt ( 2 * a * h1 )
------------------
v = sqrt ( 2 * 12 * 1000 ) = 154.919m/s
==================
g = 9.80665m/s^2
--------------------------
h2 = v^2 / (2 * g)
h = h1 + h2
h = h1 + v^2 / ( 2 * g )
h = h1 + 2 * a * h1 / ( 2 * g ) = h1 * ( 1 + a / g )
----------------------
h = 1000 * (1 + 12 / 9.80665 ) = 2223.659m
====================
t = t1 + t2
t1 = sqrt ( 2 * h1 / a )
t2 = v / g = sqrt ( 2 * a * h1 ) / g
t = sqrt ( 2 * h1 / a ) + sqrt ( 2 * a * h1 ) / g
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t = sqrt ( 2 * 1000 / 12 ) + sqrt ( 2 * 12 * 1000 ) / 9.80665 = 28.707s