An archer shoots an arrow at a 76.0 m distant target; the bull\'s-eye of the target is at same height as the release height of the arrow. (a) At what angle in degrees must the arrow be released to hit the bull\'s-eye if its initial speed is 40.0 m/s? (b) There is a large tree halfway between the archer and the target with an overhanging branch 4.93 m above the release height of the arrow. Will the arrow go over or under the branch? Solution Here, R = 76 m a) let the angle is theta R = u^2 * sin(2theta)/g 76 = 40^2 * sin(2 * theta)/9.8 solving for theta = 13.9 degree the angle is 13.9 degree b) for the maximum height h = u^2 * sin(2 *theta)/2g h = 40^2 * sin( 13.9 degree)^2/(2 * 9.8) solving for h h = 4.701 m hence , the arrow will go under the branch .