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Problem For a projectile lunched with an initial velocity of vo at an angle of e (between 0 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(0) Vinitial-y = vosin(e) а) 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, = Vịnitial-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 = Vịnitial-yt + (1/2)ayt2 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)ayt2 substituting, the vịnitial-y expression above, results to the following hmax = t+ (1/2)ayt2 Then, substituting the time, results to the following hmax = ( ) + (1/2)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 = Vinitial-xt 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 2 x ( Re-arranging and then applying the trigonometric identity sin(2x) = 2sin(x)cOS(x) we arrive at the expression for the range R as R = sin

Problem For a projectile lunched with an initial velocity of vo at an angle of e (between 0 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(0) Vinitial-y = vosin(e) а) 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, = Vịnitial-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 = Vịnitial-yt + (1/2)ayt2 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)ayt2 substituting, the vịnitial-y expression above, results to the following hmax = t+ (1/2)ayt2 Then, substituting the time, results to the following hmax = ( ) + (1/2)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 = Vinitial-xt 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 2 x ( Re-arranging and then applying the trigonometric identity sin(2x) = 2sin(x)cOS(x) we arrive at the expression for the range R as R = sin

University Physics Volume 1
University Physics Volume 1
18th Edition
ISBN:9781938168277
Author:William Moebs, Samuel J. Ling, Jeff Sanny
Publisher:William Moebs, Samuel J. Ling, Jeff Sanny
Chapter4: Motion In Two And Three Dimensions
Section: Chapter Questions
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Problem For a projectile lunched with an initial velocity of vo at an angle ofe (between 0 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 = Vịnitial-yt + (1/2)ayt2 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)ayt substituting, the vinitial-y expression above, results to the following hmax = t+ (1/2)ayt2 Then, substituting the time, results to the following hmax = ( ) + (1/2)ay( 2 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 = Vinitial-xt 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 2 x ( Re-arranging and then applying the trigonometric identity sin(2x) = 2sin(x)cos(x) we arrive at the expression for the range R as R = sin
Transcribed Image Text:Problem For a projectile lunched with an initial velocity of vo at an angle ofe (between 0 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 = Vịnitial-yt + (1/2)ayt2 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)ayt substituting, the vinitial-y expression above, results to the following hmax = t+ (1/2)ayt2 Then, substituting the time, results to the following hmax = ( ) + (1/2)ay( 2 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 = Vinitial-xt 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 2 x ( Re-arranging and then applying the trigonometric identity sin(2x) = 2sin(x)cos(x) we arrive at the expression for the range R as R = sin
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