For an "exact" solution, we can also use the following variant of the argument of Henning Makholm.
Let the height above the window be $h$, let the total time to reach the top of the window be $t_1$, and total time to reach the bottom of the window be $t_2$. Let acceleration due to gravity be constant at $a=9.8$.
Then
$$h=\frac{1}{2}at_1^2 \qquad\text{and}\qquad h+1.35=\frac{1}{2}at_2^2.$$
Subtract. We get
$$\frac{1}{2}a(t_2^2-t_1^2)=1.35.$$
But $t_2^2-t_1^2=(t_2-t_1)(t_2+t_1)$. Since $t_2-t_1=0.210$ we obtain
$$t_2+t_1=\frac{(2)(1.35)}{(0.210)(9.8)}.$$
Note that $t_1=(1/2)((t_2+t_1)-(t_2-t_1))$. Use our expression for $t_2+t_1$, together with $t_2-t_1=0.210$, to find $t_1$. Now we know $t_1$, so we know $h$.
Another way: It is more pleasant to avoid numbers until the end. Let $w$ be the height of the window, and let $s$ be the amount of time it took for the object to pass the window. Let $u$ be the velocity at the top of the window, and
$v$ the velocity at the bottom. Then the average velocity at which the window was traversed is $(v+u)/2$. But it is also $w/s$, and therefore
$$\frac{v+u}{2}=\frac{w}{s}.$$
The change in velocity is $v-u$. It is also $as$. Thus
$$\frac{v-u}{2}=\frac{as}{2}.$$
From the above two equations it follows that
$$u=\frac{w}{s}-\frac{as}{2}.$$
The average velocity from the time of dropping until arriving at the window is $u/2$. The time it took is $u/a$, so the distance travelled is $u^2/2a$. It follows that the height above the window from which the object was dropped is
$$\frac{\left(\frac{w}{s}-\frac{as}{2}\right)^2}{2a}.$$
We are given the trajectory of a projectile:
$$
y = H + x\tan(\theta) - \frac{g}{2u^2}x^2(1+\tan^2(\theta)),
$$
where $H$ is the initial height, $g$ is the (positive) gravitational constant and $u$ is the initial speed. Since we are looking for the maximum range we set $y=0$ (i.e. the projectile is on the ground). If we let $L=u^2/g$, then
$$
H + x\tan(\theta) - \frac1{2L}x^2(1+\tan^2(\theta)) = 0
$$
Differentiate both sides with respect to $\theta$.
$$
\frac{dx}{d\theta}\tan(\theta) + x\;\sec^2(\theta) - \left[\frac1L x \frac{dx}{d\theta}(1+\tan^2(\theta)) + \frac{1}{2L}x^2(2\tan(\theta)\sec^2(\theta))\right] = 0
$$
Solving for $\frac{dx}{d\theta}$ yields
$$
\frac{dx}{d\theta} = \frac{x \sec^2(\theta)[\frac{x}{L}\tan(\theta)-1]}{\tan(\theta)-\frac{x}{L}(1+\tan^2(\theta))}
$$
This derivative is $0$ when $\tan(\theta) = \frac{L}{x}$ and hence this corresponds to a critical number $\theta$ for the range of the projectile. We should now show that the $x$ value it corresponds to is a maximum, but I'll just assume that's the case. It pretty obvious in the setting of the problem. Finally, we replace $\tan(\theta)$ with $\frac{L}{x}$ in the second equation from the top and solve for $x$.
$$
H + L - \frac{1}{2L}x^2 - \frac{L}2 = 0.
$$
This leads immediately to $x = \sqrt{L^2 + 2LH}$. The angle $\theta$ can now be found easily.
Best Answer
The horizontal component $v_1$ of the velocity is an unchanging $\dfrac{7.18}{2.29}$ metres per second.
Maximum height is reached after $\dfrac{2.29}{2}$ seconds. If $v_2$ is the initial vertical component of the velocity, then the vertical component of the velocity, after $t$ seconds, is $v_2-9.81t$ (for $t\le 2.29$). This vertical component of the velocity reaches $0$ at time $\frac{2.29}{2}$. It follows that $$v_2-9.81\frac{2.29}{2}=0.$$ Now we know $v_1$ and $v_2$. The initial speed is $\sqrt{v_1^2+v_2^2}$.