(a) Displacement
The law of motion for the object is
[tex]S(t) = v_0 t + \frac{1}{2}gt^2 [/tex]
where
[tex]v_0 = 30 m/s[/tex] is the initial speed
[tex]g=-9.8 m/s^2[/tex] is the gravitational acceleration, with a negative sign because it is directed downwards, while the object is initially thrown upwards.
If we use t=2.0 s, we can find what is the displacement after 2 seconds:
[tex]S(2.0 s)=(30 m/s)(2.0 s)+ \frac{1}{2}(-9.8 m/s^2)(2.0 s)^2=40.4 m [/tex]
And this value (which is the displacement) is positive, this means that the object is still above the surface.
(b) velocity
The law of the velocity is
[tex]v(t) = v_0 + gt[/tex]
and after t=2.0 s, this is equal to
[tex]v(2.0 s)=30 m/s + (-9.8 m/s^2)(2.0 s)=10.4 m/s[/tex]
And this is positive as well, so that means that the velocity is still directed upwards, and so the object is still going up.
(c) acceleration
The acceleration of the object is constant during the motion and it is [tex]g=-9.8 m/s^2[/tex], directed downwards, so the acceleration after 2.0 s is directed downwards.
