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) 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 = 0Then,= Vinitial-y + aytSubstituting the expression of vinitial-y and ay = -g, results to the following:tThus, 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)ayt2substituting, the Vinitial-y expression above, results to the followinghmax =1+ (1/2)ayı? Then, substituting the time, results to the followinghmax = ()+ (1/2)ay(Substituting 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 horizontalaxis, 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

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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)

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)

University Physics Volume 1

18th Edition

ISBN:9781938168277

Author:William Moebs, Samuel J. Ling, Jeff Sanny

Publisher:William Moebs, Samuel J. Ling, Jeff Sanny

Chapter3: Motion Along A Straight Line

Section: Chapter Questions

Problem 113CP: The position of a particle moving along the x -axis varies with time according to x(t)=5.0t24.0t3 ....

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