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) 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? Solution