Answer
$5.33$ m
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 object launched.
At any given time, the horizontal distance traveled by the object will be $x=v_{0}\cos \theta _{0}t\left.......( 2\right) $
So using (1) and (2), we get $t=\dfrac {x}{v_{0}\cos \theta _{0}}\Rightarrow H_{t}=H_{0}+x\tan \theta _{0}-\dfrac {gx^{2}}{2v^{2}_{0}\cos ^{2}\theta _{0}} $
For the first wheel,
$v_{0}=26,5\dfrac {m}{s};X=23m;\theta _{0}=53^{0};H_{0}=3m$
Substituting these values in the formula above, we get $H_{1}=23.33m$
The height of the wheel is $18$ m so,
$\Delta H=H_{1}-H_{wheel}=23.33-18=5.33m$
