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? 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 maheight hmax and the horizontal range R. b) For what value of 0 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-yThen,Vinitial-y + aytSubstituting the expression of vinitial-y and ay = -g, results to the following:Thus, the time to reach the maximum height isImax-height =We will use this time to the equationYfinal - Yinitial - Vinitial-yt + (1/2)ay2if we use the time taken to reach the maximum height, therefore, the displacement will yield the maximum height, sohmax = Vinitial-yt + (1/2)aytsubstituting, the vinitial-y expression above, results to the followinghmax =:+(1/2)ay:? Then, substituting the time, results to the followinghmax = ()+ (1/2)aylSubstituting ay = -g, results tohmax =()- (1/2)g(2simplifying 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 groundalong the horizontal axis, so, the range R can be expressed asR = Vịnitial-xSubstituting the initial velocity on the x-axis results to the followingR = (But, 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 RasR =sin

Name: For a projectile lunched with an initial velocity of v0 at an angle of
Uploaded: 2024-03-20T06:57:47.000Z
Channel: Bartleby
Description: 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? ProblemFor a