@acer Thanks for your reply! Following your suggesion for using signum, I success like this:

solutions := (solve({v1^2 = 2*g*(h - h1), 1/2*g*t^2 = h2, v1*t + 1/2*g*t^2 = h1}, {h, t, v1}) assuming (0 < g, 0 < t, 0 < h));
expanded_solutions := allvalues(solutions);
selected_solutions := (select(sol -> is(0 < signum(subs(sol, t))), expanded_solutions) assuming (0 < g, 0 < h2));

@dharr Thanks for your reply!

@Kitonum Thanks for your reply! It generates two sets solution, if I wanna select the positive "t", I wanna try two ways: 1. the expression of t has no negative sign 2. the expression of t has no implicit part. Only the first code I write run bad. About the first one, I don't know how to express "t has no negative sign". 
 

>    (1) >  (2) >  (3)

Download select_the_positive_time_.mw

@dharr Thanks for your reply! Following your suggestion, if I want to be more rigorous, I should add "t"

ans := solve({v1^2 = 2*g*(h - h1), 1/2*g*t^2 = h2, v1*t + 1/2*g*t^2 = h1}, {g, h, t, v1});
simplify(ans);

@nm But when I set the axis range in matlab, its figure will not display the fragmented details like Maple output as below, even shrink to nothing. I wanna figure out why Maple can picture "fragmented details". 

The added range is as below

% Define the function for implicit plotting

f = @(x, y, z) x.^(2*z) + y.^(-2*y.^(-z)) + exp(-0.1*z.^2) - 1;

% Use fimplicit3 for 3D implicit plotting

fimplicit3(f, [-2 2 -2 2 -2 2]);

% Set labels and title

xlabel('x');

ylabel('y');

zlabel('z');

title('Implicit 3D Plot of f(x, y, z) = 1');

xlim([-10 -5]);

ylim([-0.5 0.5]);

zlim([-10 -5]);

@mmcdara Thanks for your reply! yes, it is exactly I want! And I try to add loop first.

@dharr Thanks for your detailed answer! it does help a lot for such cube drawinf issue

@dharr Based on the code you provided, what kind of code can make the surface of the cube have no color and its 8 edges black?

@mmcdara Thanks for your reply first! I have not try PartialSums. But i don't know how to achieve my idea with this command. My idea is similar to cutting and copying the first term (+b^2+...+k^2) to the end of the last term a^2, without losing the '+'. This can be achieved through a loop from 1 to nops(p1)/2.

@Thomas Richard Thanks for your reply!

@dharr Thanks for your reply and sorry for late reply! I will continue your code and then reply you.

@Ronan Thanks for your reply! 

Hi, I am a little interested in your original problem after reviewing your code, would you mind sharing it? I hope I didn't offend you in case it's related to your paper or something else that is not convenient to be made public.

@vv Thanks for your mention and reply!

@acer Thanks for your valuable and detailed answers！I have corrected it yesterady but still appreciate you for explaining what goes wrong.