Physics
posted by John on .
sorry to repost (again) but I'm not understanding how to solve this:
A rocket is launched at an angle of 42◦ above the horizontal with an initial speed of 76 m/s. It moves for 9 s along its initial line of motion with an acceleration of 29 m/s^2. At this time
its engines fail and the rocket proceeds to move as a projectile.
The acceleration of gravity is 9.8 m/s^2.
a.) Find the maximum altitude reached by the rocket.
b.) What is its total time of flight?
c.) What is its horizontal range?
I know how to find these with constant acceleration (I think) but with the increasing acceleration I'm not sure.
Answers are:
a.) 3837.91 m
b.) 59.9965 s
c.) 14152.7 m
A tutor told me earlier I have to break this down into 2 parts: with increasing acceleration (engines on) and after (engines off) but I'm not sure how to do the first (engines on) part.
Thank you
Need to know how to solve, thank you.

double post, sorry.

Acceleration is not increasing with engines on. It is constant.
When the engines are on,
Y = 76 sin42 t + (1/2)(29 sin 42)t^2
X = 76 cos42 t + (1/2)(29 cos 42)t^2
Use t = 9 s to find out the location and speed at burnoout. Then write a new equation for the position after that. Assume the earth is flat and that g is constant. For a rocket designed to launch satellites or escape the earth, these are only approximations.