Ray Vickson said:
Please use parentheses: you wrote ##v_2 - \frac{v_1}{t_2} - t_1##, but (I hope) you meant ##\frac{v_2-v_1}{t_2-t_1}##, which you would write in plain text as (v2-v1)/(t2-t1). You need brackets because when parsing mathematical expressions, multiplication and division take precedence over addition and subtraction, so when you write v2 - v1/t2 - t1 you first compute the fraction v1/t2, then you perform subtractions. Using parentheses over-rides those priorities and gives you what you really want.
Yes that's how i ment (v2-v1) / (t2-t1).
Ray Vickson said:
Anyway, your equation ##d = 3 t + \frac{1}{2} 2.5 \, t^2## has the acceleration pointing in the same direction as the velocity. Therefore, your ball will keep increasing its upward speed without limit, and so will blast off into outer space just like a rocket. Your ball will not reach a maximum height, it will just keep going up forever.
If I do d=3*0.4+1/2*-2.5*0.4^2
where the acceleration now is negative instead, i get correct answer (1).
d=3*2.4+1/2*-2.5*2.4^2
however, when i try to use 2.4seconds instead, i get the answer 0 if i use -2.5 acceleration, and answer 14.4 if i use +2.5 acceleration, answer is supposed to be 3.6
d=3*1.2+1/2*-2.5*1.2^2
If i use 1.2 seconds for instance, then i see the ball travels 1.8meter, and knowing there is no friction in air or the surface, i can tell that for 2.4 seconds the distance traveled must have doubled from 1.8 meter (right)?
If the above is true, does it mean that the reason i get the answer 0 from d=3*2.4+1/2*-2.5*2.4^2 is because the ball travels 1.2meter in let's say this direction ->
Then when it turns, it falls back <- to its original state, so that it says 0, where the distance from start to finish is 0.
Im not sure if this is how it works, just thinking out loud
