Please show work and diagram An archer shoots an arrow at a 72.0 m distant targe
ID: 1392549 • Letter: P
Question
Please show work and diagram
An archer shoots an arrow at a 72.0 m distant target, the bull's-eye of which is at same height as the release height of the arrow.
(a) At what angle must the arrow be released to hit the bull's-eye if its initial speed is 33.0 m/s? (Although neglected here, the atmosphere provides significant lift to real arrows.)
(b) There is a large tree halfway between the archer and the target with an overhanging branch 3.50 m above the release height of the arrow. Will the arrow go over or under the branch?
Explanation / Answer
along horizantal
initial velocity vox = vo*costheta
let T be the time total time taken by the arrow to travel to target
horizantal distance travelled ,x = vox*T =========>T = x/vox
given x = 72 m
acceleration ax = 0
along verticle
initial velocity voy = vo*sintheta
acceleration ay = -g = -9.8 m/s^2
after reaching the target the displacememnt in vertical
direction dy = 0
dy = voy*T + 0.5*g*T^2
0 = voy*T - 0.5*g*T^2
T = 2*voy/g
but T = x/vox
x/vox = 2*voy/g
g*x = 2*voy*vox
2*vo*costheta*sintheta = g*x
vo^2*sin(2*theta) = g*x
sin(2*theta) = (9.8*72)/(33*33)
theta = 20.19 degrees
++++++++++++++
part b)
at x = 72/2 = 36 m
yf = yo + 3.5
dy = 3.5 m
time taken to reach 36 m horizantally
t = x/vox = 36/(33*cos20.19) = 1.16 s
in t the vertical distance travelled by the arrow is
y = voy*t + 0.5*g*t^2
y = (33*sin20.19*1.16)-(0.5*9.8*1.6*1.6)
y = 0.667 m
y < 3.5 m
the arrow will go under the branch