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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(e) Vinitial-y = vosin(0) a) Let us first find the time 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: t 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)ayt? 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)ay? Then, substituting the time, results to the following hmax = ( )+ (1/2)ay(

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(e) Vinitial-y = vosin(0) a) Let us first find the time 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: t 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)ayt? 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)ay? Then, substituting the time, results to the following hmax = ( )+ (1/2)ay(

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Substituting ay = -g, results to hmax = ( )- (1/2)g( simplifying the expression, yields hmax = x sin 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 Ras R = sin
Transcribed Image Text:Substituting ay = -g, results to hmax = ( )- (1/2)g( simplifying the expression, yields hmax = x sin 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 Ras R = sin
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(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)ayt? 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 vinitial-y expression above, results to the following hmax = t+ (1/2)ayt? Then, substituting the time, results to the following hmax = ( )+ (1/2)ay
Transcribed Image Text: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(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)ayt? 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 vinitial-y expression above, results to the following hmax = t+ (1/2)ayt? Then, substituting the time, results to the following hmax = ( )+ (1/2)ay
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