8.55.
Model:
Visualize:
Assume the particle model and apply the constant-acceleration kinematic equations.
Solve:
(a) Newton’s second law for the projectile is
G
−F
Fnet = − Fwind = max ⇒ ax =
x
m
where Fwind is shortened to F. For the y-motion:
( )
y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 ) ⇒ 0 m = 0 m + ( v0 sin θ ) t1 − 12 gt12 ⇒ t1 = 0 s and t1 =
2
2v0 sin θ
g
Using the above expression for t1 and defining the range as R we get from the x motion:
x1 = x0 + v0 x ( t1 − t0 ) + 12 ax ( t1 − t0 )
2
⎛ 2v sin θ
⎛ F⎞
⇒ x1 − x0 = R = v0 xt1 + 12 ⎜ − ⎟ t12 = ( v0 cosθ ) ⎜ 0
g
⎝ m⎠
⎝
=
⎞ F ⎛ 2v0 sin θ ⎞
⎟−
⎜
⎟
g
⎠ 2m ⎝
⎠
2
2v02
2v 2 F
cosθ sinθ − 0 2 sin 2 θ
g
mg
We will now maximize R as a function of θ by setting the derivative equal to 0:
dR 2v02
2 Fv02
cos 2 θ − sin 2 θ ) −
2sinθ cosθ = 0
=
(
dθ
g
mg 2
⎛ 2 Fv02 ⎞ ⎛ g ⎞
mg
⇒ cos 2 θ − sin 2 θ = cos 2θ = ⎜
sin 2θ ⇒ tan 2θ =
2 ⎟⎜
2 ⎟
F
⎝ mg ⎠⎝ 2v0 ⎠
Thus the angle for maximum range is θ = 12 tan −1 ( mg / F )
(b) We have
2
mg ( 0.50 kg ) ( 9.8 m/s )
=
= 8.167 ⇒ θ = 12 tan −1 ( 8.167 ) = 41.51°
F
0.60 N
The maximum range without air resistance is
R′ =
2v02 sin 45° cos 45° v02
=
g
g
Therefore, we can write the equation for the range R as
R = 2 R′ sin 41.51° cos 41.51° −
⇒
2F
R′ sin 2 41.51° = R′ ( 0.9926 − 0.1076 ) = 0.885R′
mg
R
R′ − R
= 0.8850 ⇒
= 1 − 0.8850 = 0.115
R′
R′
Thus R is reduced from R′ by 11.5%.
Assess: The condition for maximum range ( tan 2θ = mg / F ) means 2θ → 90° as F → 0. That is, θ = 45°
when F = 0, as is to be expected.