Prof gave us this question to practice but I do not know how to solve it. If you know please provide step by step with the answer so I can understand it. Thank you very much :)

Suppose a rocket is launched from the ground with 10 seconds worth of fuel. The rocket has an upward acceleration of 8 m/s^2 while it still has fuel but after the fuel runs out, it has an acceleration of −9.8 m/s^2.

(a) Find functions describing the velocity and position of the rocket while it still has fuel.
(b) Find the velocity and height of the rocket at the moment it runs out of fuel.
(c) Find functions describing the velocity and position of the rocket after it has run
out of fuel.
(d) Find the time when the maximum height is reached by the rocket. What is the maximum height (round to one decimal place)?

4 answers

under power,

a(t) = 8
v(t) = 8t
h(t) = 4t^2
So,
v(10) = 80
h(10) = 400

Now, in ballistic path, only gravity affects the motion, so for t>10,

h(t) = 400 + 80(t-10) - 9.8(t-10)^2
= -1380 + 276t - 9.8t^2

Now you can find the vertex, the roots, etc.
hahahahahahah UW Math 127 assignment
lolimsodead
Bonjour....