QUESTION 4ProblemFor 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 thehorizontal range R. b) For what value of 0 gives the highest maximum height?SolutionThe components ofvo are expressed as follows:Vinitial-x = Vocos(0)Vinitial-y = vosin(0)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,= Vịnitial-y + aytSubstituting the expression of Vinitial-y and ay = -g, results to the following:%3DThus, the time to reach the maximum height istmax-heightWe will use this time to the equationYfinal - Yinitial = Vịnitial-yt +(1/2)ayt?%3Dif we use the time taken to reach the maximum height, therefore, the displacement will yield the maximum height, sohmax = Vinitial-yt +(1/2)ayt?%3Dsubstituting, the vinitial-y expression above, results to the followinghmax =t+ (1/2)ayt2Then, substituting the time, results to the following Then, substituting the time, results to the followinghmax = ()+ (1/2)ay%3DSubstituting ay = -g, results tohmax = () - (1/2)g(simplifying the expression, yieldshmax =x sin%3Db)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 asR = Vinitial-xtSubstituting the initial velocity on the x-axis results to the followingR = ()tBut, 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 identitysin(2x) = 2sin(x)cos(x)we arrive at the expression for the range R asR =sin

Name: QUESTION 4ProblemFor a projectile lunched with an initial velocity of
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Description: QUESTION 4ProblemFor 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 thehorizontal range R. b) For what value of 0 gives the highest maximum heigh

vfinal-y=vinitial-y+aytvfinal-y=00=vinitial-y+ayt0=vinitial-y-gttmax-height=vinitial-ygyfinal-yiniti…

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QUESTION 4 Problem For a projectile lunched with an initial velocity of vo at an angle of e (between 0 and 900), a) derive the general expression for maximum height hmax and the horizontal range R. b) For what value of 0 gives the highest maximum height? Solution The components of vo are expressed as follows: Vinitial-x = Vocos(0) Vinitial-y = vosin(0) 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 Vịnitial-y expression above, results to the following hmax = t+ (1/2)ayt? Then, substituting the time, results to the following

QUESTION 4 Problem For a projectile lunched with an initial velocity of vo at an angle of e (between 0 and 900), a) derive the general expression for maximum height hmax and the horizontal range R. b) For what value of 0 gives the highest maximum height? Solution The components of vo are expressed as follows: Vinitial-x = Vocos(0) Vinitial-y = vosin(0) 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 Vịnitial-y expression above, results to the following hmax = t+ (1/2)ayt? Then, substituting the time, results to the following

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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