1. The problem statement, all variables and given/known data

From People's Physics Book, ch. 3:

2. Relevant equations

For a. I got 23 m/s which is correct and b, 3.6 s which is correct as well. For c I chose the equation

x(t)=x0+v0t+1/2at^2

3. The attempt at a solution

And plugged things in like so... (where upwards is positive)
x(t)=20m+12m/s*t+1/2(9.8m/s^2 )(t^2 )

And then I just started in plugging in times. 1 second was the highest x(t) value I could find, at 27.1 meters, but the book says the correct answer (max height) is 28 m. Am I missing something?

Help with d would also be helpful.

Additional question: Is there a resource for the people's physics book that has explanations for problems?

Pugging random times probably won't get you to an exact answer.

The moment when the man is at the greatest height is when his velocity reaches 0. After he lets go of the rope, he keeps going up, because of the initial speed of 12m/s, after a time t=12/9.8=1.22s he stops going up ans starts to fall, this is the moment of greatest height and is the time you should put into the equation to get the answer.

I'm quite sure this is what you should do, however, i don't get the 28m the book says.

Actually, if you look at it more carefully and try to find a time for wich the height is 28m, there won't be a solution so, either the equation is not well written o the book is wrong. My guess is the book is wrong.

As for the letter d all you should do is to put t=3s in the equation you wrote. I think the answer is 8.1m.

For part c: It's as though he starts at an initial height of 20 m, with an initial upwards velocity of v_{0} = 12 m/s. Now, when he reaches his max height, his vertical velocity will have been reduced to 0 m/s. Think about it: the maximum height must be the point at which he momentarily comes to a stop. If he didn't come to a stop, then he would still be moving upwards, and so that wouldn't be his maximum height!

So, one approach that you can use here is to use the kinematics equation that tells you the velocity versus time:

Δv = at

v - v_{0} = at = -gt

Set v = 0, solve for t, and then plug this into your expression for the distance vs. time to find the height at that instant.

However, a much easier method is to use conservation of energy, if you know about that.

For d.
Find displacement of the man for 3secs after detachment.
Take point of detachment as origin. Upward positive.
Find displacement of the helicopter too.
Add the magnitude of these displacements.