a) Let us first find the time it takes for the projectile to reach the maximum...

Transcribed Image Text:

a) Let us first find the time it takes for the projectile to reach the maximum height. Using: since the y-axis velocity of the projectile at the maximum height is Then, Substituting the expression of Vinitial-y and ay = -g, results to the following: Thus, the time to reach the maximum height is We will use this time to the equation tmax-height= Then, substituting the time, results to the following hmax = ( Substituting ay = -g, results to hmax = ( simplifying the expression, yields hmax = Vfinal-y= substituting, the Vinitial-y expression above, results to the following hmax = Vfinal-y Vinitial-y + ayt Yfinal - Yinitial = Vinitial-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 x sin = Vinitial-y + ayt t+ (1/2)ayt² ) + (1/2)ay( )-(1/2)g( )/ t 1² ² a) Let us first find the time it takes for the projectile to reach the maximum height. Using: since the y-axis velocity of the projectile at the maximum height is Then, Substituting the expression of Vinitial-y and ay = -g, results to the following: Thus, the time to reach the maximum height is We will use this time to the equation tmax-height= Then, substituting the time, results to the following hmax = ( Substituting ay = -g, results to hmax = ( simplifying the expression, yields hmax = Vfinal-y= substituting, the Vinitial-y expression above, results to the following hmax = Vfinal-y Vinitial-y + ayt Yfinal - Yinitial = Vinitial-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 x sin = Vinitial-y + ayt t+ (1/2)ayt² ) + (1/2)ay( )-(1/2)g( )/ t 1² ²

Answer rating: 100% (QA)

1 a xcomponent of initial velocity VinitioV cosa Ycomponent of initi... View the full answer