A stone is thrown to hit a small target, which is at a distance of $14$ m horizontally and $5.5$ m vertically upwards from the point of projection. The stone is thrown from a height of $2$ m above the horizontal ground with a speed of $42$ m/s. Find, in degrees correct to one decimal place, the angle from the horizontal at which the stone should be thrown to hit the target. (Assume that there is no air resistance.)
Textbook Answer: $16.3^\circ$ or $87.8^\circ$
My Answer: $23.7^\circ$ or $87.7^\circ$
Does my textbook have the correct answer or is my answer correct?
I figured out the issue. You took the question to mean the target is $14$ metres horizontally and $5.5$ metres vertically from the exact point of release.
The textbook answer is based on the target being $14$ metres horizontally and $5.5$ metres vertically from the point on the ground directly (vertically) below the point of release. In other words, $14$ metres horizontally and only $3.5$ metres vertically from the exact point of release.
I got both sets of answers using the two different interpretations.
For what it's worth, I agree with your interpretation and your answers.
You can generate both sets of answers by solving this quadratic:
$0.545\tau^2 - 14\tau + (H+0.545) = 0$
where $\tau = \tan \theta, \ \theta$ being the required angle of projection from the horizontal, and $H$ being either $5.5$ or $3.5$ based on your interpretation. I took $g = 9.81$ meters per second squared.