Answer
$7.85m$
Work Step by Step
At any given time, the height of the particle will be:
$H_{t}=H_{0}+v_{0}\sin \theta _{0}t-\dfrac {gt^{2}}{2}\left.....( 1\right) $
where $H_{0}$ is initial height of particle, $v_{0}$ is initial velocity, $\theta_{0}$ is the initial angle of particle relative to horizontal line and $t$ is the time passed since the object launched.
Lets calculate the maximum height of the object:
$H_{\max }=H_{0}+\dfrac {v^{2}_{0}\sin ^{2}\theta _{0}}{2g}=3m+\dfrac {\left( 26.5\dfrac {m}{s}\right) ^{2}\sin ^{2}53^{\circ}}{2\times 9.8\dfrac {m}{s^{2}}}=25.85m$
Therefore,
$\Delta H=H_{\max -}H_{wheel}=25.85-18=7.85m$
