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Get Full Access to Physics: Principles With Applications - 6 Edition - Chapter 1 - Problem 60gp
Get Full Access to Physics: Principles With Applications - 6 Edition - Chapter 1 - Problem 60gp

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# A large passenger aircraft accelerates down the runway for

ISBN: 9780130606204 3

## Solution for problem 60GP Chapter 1

Physics: Principles with Applications | 6th Edition

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Physics: Principles with Applications | 6th Edition

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Problem 60GP

A large passenger aircraft accelerates down the runway for a distance of 3000 m before leaving the ground. It then climbs at a steady 3.0° angle. After the plane has traveled 3000 m along this new trajectory, (a) how high is it, and (b) how far horizontally is it, from its initial position?

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ANSWER: STEP 1:- Before leaving the ground the aircraft ran 3000 m on the ground. Then it got lifted with the angle of 3.0° and continued for 3000 m along the inclined trajectory. So schematically the problem looks like the figure given below. In the above diagram, A is the starting point of the aircraft. AB is the distance travelled before leaving the ground = 3000 m, 3° is the angle the aircraft made with the horizontal after leaving the ground, CD is the height the aircraft reached. BC is the distance covered through the new path after take off. STEP 2:- a) how high is it In the diagram, triangle BCD looks like a right angled triangle, whose hypotenuse and the angle we know. So by applying the laws of trigonometry, Let’s use the relation of, sin = perpendicular / hypotenuse 0 sin 3 = CD / BC 0.052 = CD / 3000 CD = 0.052 × 3000 = 157.007 = 157 m(approximately) But CD is the height. So the aircraft is at a height of 157 meters. c) how far horizontally is it, from its initial position To answer this question, we will use the famous pythagoras theorem for the triangle BCD. Here we know the height 157 m and the hypotenuse 3000 m. to get the base which is the required quantity, h = p + b -----------------------------------------(1) base = h p = 3000 157 = 8975351 = 2995.88 m So, So the horizontal displacement is 2996 m approximately. STEP 3:- But we are interested in the horizontal displacement from it’s initial position A to the final horizontal position D. So, net horizontal displacement = AB+BD = 3000+2996 = 5996 m. CONCLUSION: The aircraft is 5996 meters far from it’s initial position horizontally.

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