A ground to ground projectile is at point t=T/3 is at point t=5T/6 and reaches ground at t=T the difference in heights between B and A is
A- \[\dfrac{g{{T}^{2}}}{6}\]
B- \[\dfrac{g{{T}^{2}}}{12}\]
C- \[\dfrac{g{{T}^{2}}}{18}\]
D- \[\dfrac{g{{T}^{2}}}{24}\]

Hint: The equation of projectile is,$y=x\tan \theta -\dfrac{g{{x}^{2}}}{2{{u}^{2}}{{\cos }^{2}}\theta }$ where $u$is velocity of projection and $\theta $ is the angle of projection. Projectile moves under the influence of gravitational force with no acceleration in horizontal direction.

Complete step by step answer:Let the difference between A and B = H
As time is given by the formula, \[T=\dfrac{2u\sin \theta }{g}\]
\[\dfrac{gT}{2}=u\sin \theta \]--(1)
Now for A: t=T/3
\[\begin{align}
  & s={{u}_{y}}t+\dfrac{a{{t}^{2}}}{2} \\
 & h=u\sin \theta t-\dfrac{g{{t}^{2}}}{2} \\
\end{align}\]
Now substituting the value of t, for case A we get,
\[\begin{align}
  & h=\dfrac{gT}{2}\times \dfrac{T}{3}-\dfrac{g{{T}^{2}}}{18} \\
 \Rightarrow & h=\dfrac{g{{T}^{2}}}{6}-\dfrac{g{{T}^{2}}}{18} \\
\end{align}\]
\[h=\dfrac{2g{{T}^{2}}}{18}\]---(2)
Now for A: t=5T/6
\[\begin{align}
  & s={{u}_{y}}t+\dfrac{a{{t}^{2}}}{2} \\
  \Rightarrow & h'=u\sin \theta t-\dfrac{g{{t}^{2}}}{2} \\
\end{align}\]
Now substituting the value of t, for case B we get,
\[\begin{align}
  & h'=\dfrac{gT}{2}\times \dfrac{5T}{6}-\dfrac{25g{{T}^{2}}}{72} \\
  \Rightarrow & h'=\dfrac{5g{{T}^{2}}}{12}-\dfrac{25g{{T}^{2}}}{72} \\
\end{align}\]
\[h'=\dfrac{5g{{T}^{2}}}{72}\]--(3)
Now we have assumed the difference between A and B to be H.
\[\therefore H=h-h'\]
\[\begin{align}
   \Rightarrow & \dfrac{2g{{T}^{2}}}{18}-\dfrac{5g{{T}^{2}}}{72} \\
  \Rightarrow & \dfrac{3g{{T}^{2}}}{72} \\
\end{align}\]
\[\therefore H=\dfrac{g{{T}^{2}}}{24}\]
So, the correct option is (D)

Additional information- In projectile problems when the body is projected whether, from the ground or a certain height, there is no acceleration in the horizontal direction. There is constant acceleration in the vertical direction which is always directed downwards and this is constant. This is the acceleration due to gravity.

Note:while solving problems which include projectile, certain things are to be kept in mind such as in the formula we have to put the angle that the projectile makes with the horizontal. The motion of the body has been split into its components and that is possible only because of the fact that acceleration of the body is constant.