Refer to the pictures attached. ProblemFor 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?SolutionThe 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 + 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)ayt? if 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)ayl2Substituting ay = -g, results tohmax = () - (1/2)g(simplifying the expression, yieldshmax =x sinb)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 canbe 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

(a) The components of the initial velocity along x and y direction can be written as v0x=v0cosθ…

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 8 gives the highest maximum height?

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 8 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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Kindly fill in the blanks. 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? The components of v0 are expressed as follows: vinitial-x = v0cos(θ) vinitial-y = v0sin(θ)
Problem: 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)
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)
Problem 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…
Problem 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 isvfinal-y =___________ Then, ___________= vinitial-y + aytSubstituting the expression of vinitial-y and ay = -g, results to the following: __________=__________-__________t Thus, the time to reach the maximum height istmax-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 =…
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