help~

Hey
hey pc6817


I suspect you are getting your HOMEWORK done from us ...All questions you have posted are simple kinematics problems!

Be Honest and tell us

I don't mind helping you but you could have posted them under same thread for your and our ease

Isn't it?

For part A, you have the upward thrust of the engine and the downward pull of gravity. If you get a negative number, something is wrong.

You need to break up the problem into two different stages since you have a step change in acceleration. Solve for height and velocity at 16s, then use that as a starting point for the second stage of travel. As you've seen in other posts, the velocity at time t = Vo + At and the position at time t = (Vo)t + 1/2 A t^2 + Ho where Ho is the initial height and Vo is the initial velocity.

If A is the initial acceleration, and the rocket is initially at rest then the height after 16 sec is (1/2) A t^2 = 128 A. The velocity at that point is At = 16A

In the second stage, use the velocity calculated above for the initial velocity and the height calculated above for the initial height to solve the equation again. It's ok that those values are in terms of A. Remember to use the time AFTER THE INITIAL 16 seconds. As you noted, the new acceleration is -9.8 m^2/sec. You now have an equation in terms of the initial A that you can use that 5100 meters at 20 seconds data point in. (Remember, 20-16 seconds in the equation).

A couple of notes:
- The weight of the rocket is unimportant.
- We already know this is homework and many of us don't have any problem with that as long as you are trying to solve the problems on your own.