(a) Explanation Given: The projectileis travelling horizontally at 20 m/s and its position is relative to the origin ( 0 , 0 ) . The trajectory of the path followed by packet is x = ( 20 cos θ ) t , y = − 4.9 t 2 + ( 20 sin θ ) ; t ≥ 0 . Calculation: Substitute θ = π 6 in x = ( 20 cos θ ) t , y = − 4.9 t 2 + ( 20 sin θ ) and obtain the values of x and y as shown below. x = ( 20 cos π 6 ) t = ( 10 3 ) t y = − 4.9 t 2 + ( 20 sin π 6 ) t = − 4.9 t 2 + 10 t Use online graphing calculator and draw the graph of x = 10 3 t , y = − 4.9 t 2 + 10 t as shown below in Figure 1. From the Figure 1 it can be noted that the trajectory followed by the projectile is parabolic. Substitute θ = π 4 in x = ( 20 cos θ ) t , y = − 4.9 t 2 + ( 20 sin θ ) and obtain the values of x and y as shown below. x = ( 20 cos π 4 ) t = ( 10 2 ) t y = − 4 (b) To determine To observe: the value of θ that gives the largest distance between the point of launching and point of landing.

2nd Edition

ISBN: 9781323178522

Author: Briggs

Publisher: PEARSON

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Question

Chapter 10.1, Problem 106E

(a)

To determine

To graph: The trajectory of a projectile launched from the ground for various values of θ.

(b)

To determine

To observe: the value of θ that gives the largest distance between the point of launching and point of landing.

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Chapter 10.1, Problem 105E

Chapter 10.1, Problem 107E

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