fisica1-youn-e-freedman-exercicios-resolvidos

3.73: a) The acceleration is given as g at an angle of 53 .1°
to the horizontal. This is a 3-
4-5 triangle, and thus, a = ( 3/5) g and a = (4/5 g during the "boost" phase of the flight.
x y
)
Hence this portion of the flight is a straight line at an angle of 53 .1°
to the horizontal.
After time T, the rocket is in free flight, the acceleration is a = 0 and a y
= g , and the
familiar equations of projectile motion apply. During this coasting phase of the flight, the
trajectory is the familiar parabola.
x
b) During the boost phase, the velocities are: v x
= (3/5)
gt and v y
= (4 /5)
gt , both
straight lines. After t = T , the velocities are v x
= (3/5)
gT , a horizontal line, and
v y
= ( 4/5) gT − g(
t − T ) , a negatively sloping line which crosses the axis at the time of
the maximum height.
c) To find the maximum height of the rocket, set v = 0 , and solve for t, where t = 0
when the engines are cut off, use this time in the familiar equation for y. Thus, using
t = (4/5)T
and
1 2
2
2
2
2 8
y , 2
4 (
4 )
1 (
4 ) ,
2
16
max
= y0
+ v0
yt
−
2
gt ymax
=
5
gT +
5
gT
5
T −
2
g
5
T ymax
=
5
gT +
25
gT −
25
gT
18 2
Combining terms, y
max
= gT
25
.
d) To find the total horizontal distance, break the problem into three parts: The boost
phase, the rise to maximum, and the fall back to earth. The fall time back to earth can be
2
2
found from the answer to part (c), ( 18/ 25) gT = (1/ 2) gt , or t = (6/5)
T . Then,
2
6
multiplying these times and the velocity, x = 3 gT +
3 gT )(
4 T ) + (
3 gT )( T ) , or
10
2 12 2 18 2
= 3 gT + gT gT
10 25 25
. Combining terms gives
x +
y
(
5 5 5 5
3 2
gT
2
x = .