fisica1-youn-e-freedman-exercicios-resolvidos

3.74: In the frame of the hero, the range of the object must be the initial separation plus
the amount the enemy has pulled away in that time. Symbolically,
R
R = x0 + vE/Ht
= x0
+ vE/H
, where v
E/H
is the velocity of the enemy relative to the hero, t
v0
x
is the time of flight, v
0x
is the (constant) x-component of the grenade’s velocity, as
measured by the hero, and R is the range of the grenade, also as measured by the hero.
Using Eq. (3-29) for R, with sin 2 α = 0
1 and v / 2
x
= ,
0
v 0
2
v0
v0
2
= x0
+ vE/H
2,
or v0
− ( 2vE/H
) v0
− gx0
= 0.
g
g
This quadratic is solved for
1
2
v
0
= ( 2vE/H
+ 2vE/H
+ 4gx0
) = 61.1km/h,
2
where the units for g and x
0
have been properly converted. Relative to the earth, the x-
component of velocity is 90 .0 km/h + (61.1km/h)cos45°
= 133.2 km/h , the y-component,
the same in both frames, is ( 61.1km/h)sin 45° = 43.2 km/h , and the magnitude of the
velocity is then 140 km/h.
2 2
2
2 2 2
2
2
3.75: a) x + y = ( Rcosωt)
+ ( Rsin
ωt)
= R (cos ω t + sin ωt)
= R ,
so the radius is R.
b) v = −ω Rsinωt,
v = ωRcosωt,
and so the dot product
r ⋅ v
r = xv
x
= ωR(
− cosωt
sinωt
+ sinωt
cosωt)
= 0.
x
+ yv
y
y
= ( R cosωt)(
−ωR
sinωt)
+ ( R sinωt)(
ωR
cosωt)
2
2
2
2
c) ax
= −ω
R cosωt
= −ω
x, ay
= ω Rsinωt
= −ω
y,
r 2 r
2
and so a = −ω and a = ω R.
d) v
v = ωR .
e)
ω ω ω ω ω ω ω = ω
2 2 2
2
2 2 2 2
2
2 2
= vx
+ vy
= ( − Rsin
t)
+ ( Rcos
t)
= R (sin t + cos t)
R , and so
( ωR)
2
2
v
a = ω R = = .
R R
2