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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 Imax-height We will use this time to the equation yfinal - Yinitial = Vinitialyt+ (1/2)a,2 if we use the time taken to reach the maximum height, therefore, the displacement will yield the maximum height, so hmax = Vinitial-y + (1/2lay? substituting the Vinitial-y expression above, results to the following hmax = Then, substituting the time, results to the following hmax =( + (1/2lay Substituting ay =-g results to hmax =( )- (1/2ig( 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=vinitial-x Substituting the initial velocity on the x-axis results to the following R=( 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 2 x( Re-arranging and then applying the trigonometric identity sin(2x) = 2sin(xlcos(x) we arrive at the expression for the range Ras R= sin

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 Imax-height We will use this time to the equation yfinal - Yinitial = Vinitialyt+ (1/2)a,2 if we use the time taken to reach the maximum height, therefore, the displacement will yield the maximum height, so hmax = Vinitial-y + (1/2lay? substituting the Vinitial-y expression above, results to the following hmax = Then, substituting the time, results to the following hmax =( + (1/2lay Substituting ay =-g results to hmax =( )- (1/2ig( 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=vinitial-x Substituting the initial velocity on the x-axis results to the following R=( 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 2 x( Re-arranging and then applying the trigonometric identity sin(2x) = 2sin(xlcos(x) we arrive at the expression for the range Ras R= sin

Classical Dynamics of Particles and Systems
Classical Dynamics of Particles and Systems
5th Edition
ISBN:9780534408961
Author:Stephen T. Thornton, Jerry B. Marion
Publisher:Stephen T. Thornton, Jerry B. Marion
Chapter2: Newtonian Mechanics-single Particle
Section: Chapter Questions
Problem 2.6P: In the blizzard of ’88, a rancher was forced to drop hay bales from an airplane to feed her cattle....
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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 Imax-height We will use this time to the equation yfinal - Yinitial = Vinitialyt+ (1/2)a,2 if we use the time taken to reach the maximum height, therefore, the displacement will yield the maximum height, so hmax = Vinitial-y + (1/2lay? substituting the Vinitial-y expression above, results to the following hmax = Then, substituting the time, results to the following hmax =( + (1/2lay Substituting ay =-g results to hmax =( )- (1/2ig( 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=vinitial-x Substituting the initial velocity on the x-axis results to the following R=( 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 2 x( Re-arranging and then applying the trigonometric identity sin(2x) = 2sin(xlcos(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 Imax-height We will use this time to the equation yfinal - Yinitial = Vinitialyt+ (1/2)a,2 if we use the time taken to reach the maximum height, therefore, the displacement will yield the maximum height, so hmax = Vinitial-y + (1/2lay? substituting the Vinitial-y expression above, results to the following hmax = Then, substituting the time, results to the following hmax =( + (1/2lay Substituting ay =-g results to hmax =( )- (1/2ig( 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=vinitial-x Substituting the initial velocity on the x-axis results to the following R=( 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 2 x( Re-arranging and then applying the trigonometric identity sin(2x) = 2sin(xlcos(x) we arrive at the expression for the range Ras R= sin
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