# Stuck On A Projectile Motion Question, Is My Answer Correct Or The Textbook Answer?

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?

• In case you are wondering, I tried to edit the question to display the image directly in the question. Something misfired, so I deleted my editing. Commented Apr 16, 2021 at 3:09
• Welcome to MSE. Don't use picture in the question session. Use MathJax formatting to mathematical expressions. See math.meta.stackexchange.com/questions/5020/… Commented Apr 16, 2021 at 3:15
• @user2661923 That imgur address has two images so you need to separately include them when inlining. Commented Apr 16, 2021 at 3:16
• @user10354138 Thanks, the situation had me scratching my head. I wasn't seeing the *.jpg links. Commented Apr 16, 2021 at 3:41

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.

$$0.545\tau^2 - 14\tau + (H+0.545) = 0$$
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.