For a projectile lunched with an initial velocity of v0 at an angle of θ (between 0 and 90o) . 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) 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, therange 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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Description: For a projectile lunched with an initial velocity of v0 at an angle of θ (between 0 and 90o) . 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) The

Solution:) Given that Velocity is v0 The components of v0 are expressed as follows: vinitial-x =…

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For a projectile lunched with an initial velocity of v0 at an angle of θ (between 0 and 90o) . 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) . 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)

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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