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