Kinematics M1

I found a question in my M1 book,

A particle is projected vertically upwards with a speed of 30m/s-1 from point A. The point B is h metres above A. The particle moves freely under gravity and its above A for a time 2.4s. Find the value of h

What do they actually mean by moves freely under gravity? That gravity does not apply to it, and in this case it would mean that I certainly do not have enough information, and if it means that it still has the acceleration of -9.8m/s-2 I still do not get the right answer

Here are my workings :

u = 30
a = -9.8
v= 0

Via v = u + at, I get that t = 3.06s, therefore it is in the air for a total of 3.06 seconds

It is 2.4 above B, so it is under B(until it reaches h) for 0.66 seconds

using s=ut+1/2at^2 , I get that that s = 17.66556 , and the right answer is 39 to s.f? :s

Any help ?

It moves freely under gravity means that it doesn't have any other forces acting on it. E.g. air resistance.

EDIT: Hold on, just realised something. Gimme a sec to work this out lol.

EDIT2: Ok, as far as I can remember (I did M1 2 years ago now, but I did get a high A in it), the time is irrelevant in this question. Here is my working:

s ?
u 30
v 0
a -9.8
t

900 + 2(-9.8)(s)
-900 = -19.6s
19.6s = 900
s = 900/19.6

using the formula v^2 = u^2 +2as. Since v = 0 I have omitted it.

Your method is right apart from one small mistake: 3.06 is not the total time in the air, it is the time until it reaches the maximum. When the particle goes up, it is above b for half of 2.4 seconds (1.2) before it reaches the maximum height, then another 1.2 seconds after.

So, using your method, you are looking at just the first half of motion, so you only need to take off 1.2 seconds from 3.06. Does that get you the right answer?

(This is all assuming the bolded 'A' should read B in the question, judging from your working).

You got the maximum heigth reached, exactly as mine.