Answer
The keys will land $3.61\text{ m}$ left of the base.
Work Step by Step
We will first find his height at that moment. The height with respect to the center of the wheel is $h1=R\cos\theta$ where theta is the angle between 12 o' clock and 10 o' clock so its value is $2\cdot 30^\circ=60^\circ.$ Thus, $h_1=5\cos60^\circ\text{ m}=2.5\text{ m}$. The height of the center with respect to the bottom is $h_2=R=5\text{ m}$ and the height of the bottom with respect to the base is $h_2=1.75\text{ m}$. So the total height of the passenger is $h=h_1+h_2+h_3=9.25\text{ m}$. Now, the velocity at the time of dropping points along the tangent of the wheel i.e. perpendicular to the radius at 10 o' clock so the "launch" angle is also $\theta=60^\circ$ above the horizontal (by the theorem of angles with perpendicular legs). The launch speed is the same as the speed of the rotation of the wheel at that point i.e. $v_0=\frac{2\pi R}{32\text{ s}}=0.98\text{ m/s}$.
Now putting the origin at the base and directing the $x$ axis rightwards, the equations of motion are
$$x=v_0t\cos60^\circ-5\sin60^\circ\text{ m}=\frac{1}{2}v_0t-\frac{5\sqrt{3}}{2}m$$
$$y=v_0t\sin60^\circ-\frac{1}{2}gt^2+h=\frac{\sqrt{3}}{2}v_0t-\frac{1}{2}gt^2+h.$$
The key lands at time $t$ when $y=0$ so we have
$$-\frac{1}{2}gt^2+\frac{\sqrt{3}}{2}v_0t+h=0$$
which yields
$$t=\frac{-\frac{\sqrt{3}}{2}v_0-\sqrt{\frac{3}{4}v_0^2+2gh}}{-g}=1.46\text{ s}.$$
The other solution, that is negative, has no physical meaning in this case.
Now, putting this value for $t$ in the equation for $x$ we get
$$x=\frac{1}{2}\cdot0.98\text{ m/s }\cdot 1.46\text{ s}-\frac{5\sqrt{3}}{2}\text{ m}=-3.61\text{ m},$$
i.e. the keys will land $3.61\text{ m}$ left of the base.
