Burning Bale from Catapult to Travel 150 ft and Clear Over 35 ft Castle Wall | CE Board Problem in Mathematics, Surveying and Transportation Engineering

Date of Exam

May 2016

Subject

Mathematics, Surveying and Transportation Engineering

Problem
A catapult is placed 100 ft from the castle wall, which is 35 feet high. The soldier wants the burning bale of hay to clear the top of the wall and land 50 feet inside the castle wall. If the initial velocity of the bale is 70 feet per second, then at what angle should the bale of hay be launched so that it travel 150 feet and pass over the castle wall. Use g = 32 ft/sec².

A. 49.8°	C. 39.2°
B. 50.8°	D. 40.2°

Answer Key

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[ B ]

Solution

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$y = x\tan \theta - \dfrac{gx^2}{2{v_o}^2 \cos^2 \theta}$

$0 = 150\tan \theta - \dfrac{32(150^2)}{2(70^2) \cos^2 \theta}$

$\theta = 39.2^\circ ~ \text{and} ~ 50.8^\circ$

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$y = x\tan \theta - \dfrac{gx^2}{2{v_o}^2 \cos^2 \theta}$

$0 = 150\tan \theta - \dfrac{32(150^2)}{2(70^2) \cos^2 \theta}$

$150 \tan \theta = \dfrac{3600}{49\cos^2 \theta}$

$\cos^2 \theta \, \tan \theta = \dfrac{3600}{7350}$

$\sin \theta \, \cos \theta = \dfrac{24}{49}$

$2\sin \theta \, \cos \theta = \dfrac{48}{49}$

$\sin 2\theta = \dfrac{48}{49}$

$2\theta = 78.4^\circ ~ \text{and} ~ 101.6^\circ$

$\theta = 39.2^\circ ~ \text{and} ~ 50.8^\circ$

$2016-may-math-catapult-burning-bale.gif$

Check the height at x = 100 ft for θ = 39.2°

$y = x\tan \theta - \dfrac{gx^2}{2{v_o}^2 \cos^2 \theta}$

$y = 100\tan 39.2^\circ - \dfrac{32(100^2)}{2(70^2) \cos^2 39.2^\circ}$

$y = 27.2 ~ \text{ft} \lt 35 ~ \text{ft}$ ← (not okay!)

Check the height at x = 100 ft for θ = 50.8°

$y = x\tan \theta - \dfrac{gx^2}{2{v_o}^2 \cos^2 \theta}$

$y = 100\tan 50.8^\circ - \dfrac{32(100^2)}{2(70^2) \cos^2 50.8^\circ}$

$y = 40.9 ~ \text{ft} \gt 35 ~ \text{ft}$ ← (okay!)

Use θ = 50.8° ← Answer: [ B ]