For a projectile lunched with an initial velocity of vo at an angle of 0 (between 0 and 90°), a) derive the general expression for maximum height hmax and the horizontal range R. b) For what value of 0 gives the highestmaximum height?SolutionThe components of vo are expressed as follows:Vinitial-x = Vocos(0)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+ aytsince the y-axis velocity of the projectile at the maximum height isVfinal-y =Then,= Vinitial-y+ aytSubstituting the expression of vinitial-y and ay = -g, results to the following:Thus, the time to reach the maximum height istmax-height "We will use this time to the equationyfinal - Yinitial - Vinitial-yt + (1/2)aytif we use the time taken to reach the maximum height, therefore, the displacement will yield the maximum height, sohmax = Vinitial-yt + (1/2)ayt?substituting, the vinitial-y expression above, results to the followinghmax =t+ (1/2)ayt?Then, substituting the time, results to the followinghmax =()+ (1/2)aySubstituting ay - -g, results tohmax = ()- (1/2)g(simplifying the expression, yieldshmax =x sin

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We know that the projectile reaches the high point of its trajectory (vy = 0) so that at time…

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For a projectile lunched with an initial velocity of vo at an angle of 8 (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?

For a projectile lunched with an initial velocity of vo at an angle of 8 (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?

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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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…
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For a projectile lunched with an initial velocity of v0 at an angle of θ (between 0 and 90o) , a) derive the general expression for maximum height hmax and the horizontal range R. b) For what value of θ gives the highest maximum height? Solution The components of v0 are expressed as follows: vinitial-x = v0cos(θ) vinitial-y = v0sin(θ) 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: = - 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)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 vinitial-y expression above, results to…
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