Problem: An archer shoots an arrow at a 75 m distant target; the bull’s-eye of the
target is at same height as the release height of the arrow. If d = 75 m and v = 37 m/s.
At what angle must the arrow be released to hit the bull’s-eye if its initial speed is
37m/s?
Solution:
This is projectile motion problem and by analyzing the problem, we are given
the range and the initial velocity.
range
=75
m
initial veloctiy
=37
m
/
s
To find the angle of release of the arrow to hit the bull’s-eye, we will use the
formula for range
R
=
u
2
∙
sin 2
θ
g
where
R
=
range
u
=
initial velocity
g
=
gravity acceleration
(We will use
9.81
m
s
2
for
g
)
θ
=
inital launch angle
By solving
θ
in terms of other variables, we have
gR
=
u
2
∙
sin 2
θ
Cross multiplication
sin 2
θ
=
gR
u
2
Division Property of Equality
sin−1
(
sin 2
θ
)
=sin−1
(
gR
u
2
)
Applying inverse sine to both sides of
the equation
2
θ
=sin−1
(
gR
u
2
)
θ
=1
2
∙
sin−1
(
gR
u
2
)
Multiplication Property of Equality
By substituting the values, we have
θ
=1
2
∙
sin−1
[
(
9.81
m
s
2
)
(
75
m
)
(
37
m
s
)
2
]
