Question taken from the book is in the image above.
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The solution in the back of the book is Solution $2$. However, I believe that the partcile is on the way down when $\theta = 37^\circ.$
$\tan \theta = \frac{3}{4},\ $ so the trajectory is $y = \frac{3}{4}x - \frac{5}{144}x^2$.
$\frac{dy}{dx} = \frac{3}{4} - \frac{5}{72}x.\ $ When $x=12,\ \frac{dy}{dx} = \frac{3}{4} - \frac{5 \times 12}{72} = -\frac{1}{12}< 0,\ $ so the particle is on the way down.
I then thought that the person who wrote the question wrote the solution forgetting their intended solution: for $\theta = 37^\circ\ $ to not be a possible answer because the trajectory of the ball would pass through the platform, meaning that the ball would in fact hit the underneath of the platform.
So, let's see if this is the case. When $\tan\theta = \frac{3}{4},\ $ what is $x$ when $y=4\ ?$ Well, when $y=4,\ -\frac{5}{144}x^2 + \frac{3}{4}x - 4 = 0 \implies x = 12\ $ or $\ x =9.6.\ $ But this is before the platform, and so surely $\theta = 37^\circ\ $ is another solution?