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You are on the roof of a Physics Building, 19 m above the ground. Your physics professor, who is 1.87 m tall, is approaching the building at constant speed of 2.28 m/s. If you wish to drop an egg on his head, how far should your professor be when you release the egg?

You are on the roof of a Physics Building, 19 m above the ground. Your physics professor, who is 1.87 m tall, is approaching the building at constant speed of 2.28 m/s. If you wish to drop an egg on his head, how far should your professor be when you release the egg?

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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You are on the roof of a Physics Building, 19 m above the ground. Your physics professor, who is 1.87 m tall, is approaching the building at constant speed of 2.28 m/s. If you wish to drop an egg on his head, how far should your professor be when you release the egg?
Transcribed Image Text:You are on the roof of a Physics Building, 19 m above the ground. Your physics professor, who is 1.87 m tall, is approaching the building at constant speed of 2.28 m/s. If you wish to drop an egg on his head, how far should your professor be when you release the egg?
Problem For a projectile lunched with an initial velocity of vo at an angle of e (between O and 90°). a) derive the general expression for maximum height hmax and the horizontal range R. b) For what value of e gives the highest maximum height? Solution The components of vo are expressed as follows: Vinitial-x = vocos(e) Vinitial-y= Vosin(e) a) Let us first find the time it takes for the projectile to reach the maximum height. Using: Vfinal-y = Vinitial-y + ayt since the y-axis velocity of the projectile at the maximum height is Vfinal-y= Then, = Vinitial-y * ayt Substituting the expression of Vinitial-y and ay = -g. results to the following: Thus, the time to reach the maximum height is tmax-height = We will use this time to the equation Yfinal - Yinitial = Vinitial-yt + (1/2)a, if we use the time taken to reach the maximum height, therefore, the displacement will yield the maximum height, so hmax = Vinitial-yt + (1/2)ay? substituting, the vinitial-y expression above, results to the following hmax = 1+ (1/2)a,? Then, substituting the time, results to the following hmax )+ (12)ay( = ( Substituting ay = -g, results to hmax = )- (1/2)g( simplifying the expression, yields hmax= X sin b) The distance traveled by a projectile follows a uniform motion, meaning, velocity is constant from the start point until it reach the ground along the horizontal axis, so, the range R can be expressed as R= Viritial-x Substituting the initial velocity on the x-axis results to the following R= ( )t But, the time it takes a projectile to travel this distance is just twice of tmax-height by substitution, we obtain the following: R = x 2x( Re-arranging and then applying the trigonometric identity sin(2x) = 2sin(x)cos(x) we arrive at the expression for the range Ras R = sin
Transcribed Image Text:Problem For a projectile lunched with an initial velocity of vo at an angle of e (between O and 90°). a) derive the general expression for maximum height hmax and the horizontal range R. b) For what value of e gives the highest maximum height? Solution The components of vo are expressed as follows: Vinitial-x = vocos(e) Vinitial-y= Vosin(e) a) Let us first find the time it takes for the projectile to reach the maximum height. Using: Vfinal-y = Vinitial-y + ayt since the y-axis velocity of the projectile at the maximum height is Vfinal-y= Then, = Vinitial-y * ayt Substituting the expression of Vinitial-y and ay = -g. results to the following: Thus, the time to reach the maximum height is tmax-height = We will use this time to the equation Yfinal - Yinitial = Vinitial-yt + (1/2)a, if we use the time taken to reach the maximum height, therefore, the displacement will yield the maximum height, so hmax = Vinitial-yt + (1/2)ay? substituting, the vinitial-y expression above, results to the following hmax = 1+ (1/2)a,? Then, substituting the time, results to the following hmax )+ (12)ay( = ( Substituting ay = -g, results to hmax = )- (1/2)g( simplifying the expression, yields hmax= X sin b) The distance traveled by a projectile follows a uniform motion, meaning, velocity is constant from the start point until it reach the ground along the horizontal axis, so, the range R can be expressed as R= Viritial-x Substituting the initial velocity on the x-axis results to the following R= ( )t But, the time it takes a projectile to travel this distance is just twice of tmax-height by substitution, we obtain the following: R = x 2x( Re-arranging and then applying the trigonometric identity sin(2x) = 2sin(x)cos(x) we arrive at the expression for the range Ras R = sin
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