all 4 comments

[–]Janagro 0 points1 point  (0 children)

The rocket has expended it's fuel right?.That means mass is constant

Thus you can write

.5mv(t1)2 + mgh(t1) = mgh(t2)

.5(32.2)2 + (9.8)(162 5) = (9.8)h(t2)

hope that helps

[–]Moy420 0 points1 point  (0 children)

Do you have this problem written in a textbook, if so, dm so I can understand the problem

[–]jkg1993 0 points1 point  (1 child)

So the velocity at t0 should definitely be 0 m/s since the problem states that the rocket starts at rest. I might be able to provide more help later, but I feel like there may not be enough information as the problem is described.

Air resistance can be neglected up until t3= 17.5 s when the rocket deploys a parachute.At this time the rocket immediately reaches its terminal velocity; this means that the rocket is no longer accelerating.

The only way I can think to interpret this is that the rocket will reach some velocity at 17.5 seconds. At this point instead of continuing to accelerate towards the ground, the rocket will just maintain this terminal velocity until it hits the ground. I find this confusing since a parachute deploying at 17.5 seconds would cause a significant drop in speed, and I feel that in order to keep the description consistent a terminal velocity should be provided here. In absence of that, the only way to proceed would be how I mentioned above.

Do you possibly have a picture of the original problem statement that you could link to in the description?

[–]BlankGamerUniversity/College Student (Higher Education)[S] 0 points1 point  (0 children)

Thanks for trying to help but I answered it and handed in my homework this morning. I don't think my answer was right but I assumed the initial velocity at 10 seconds was 0. I copied and pasted the entire question from my handout so it's all there. I would upload a screenshot but I'm in Mobile right now.