fisica1-youn-e-freedman-exercicios-resolvidos

3.48:
a)
The equations of motions are:
1
y = h + ( v0
sin α)
t − gt
2
x = ( v cos α)
t
v
v
y
x
= v sin α − gt
0
= v cos α
0
0
Note that the angle of 36. 9
o results in sin 36.9° = 3/5 and cos 36.9° = 4/5 .
b) At the top of the trajectory, v = 0 . Solve this for t and use in the equation for y to
find the maximum height: t
reduces to
y h +
v 2 sin 2
0 α
2g
y
v sin
sin
= . Then, ( ) ( ) 1
2
0 α v0
α
y h + v sin α)
− g
v0
sin α
g
2
= , which
(
0
g 2 g
= . Using v
0
= 25gh
/ 8 , and sin α = 3/ 5 , this becomes
(25gh
/ 5)
9
y h
8)(3 /
25
= + = h + h , or = h . Note: This answer assumes that y
0
= h . Taking
2g
2
16
y
16
y = 0 0
will give a result of y 9
= h
16
(above the roof).
c) The total time of flight can be found from the y equation by setting y = 0 , assuming
y
0
= h , solving the quadratic for t and inserting the total flight time in the x equation to
find the range. The quadratic is 1 2 3
gt − v − h = 0 . Using the quadratic formula gives
(3 / 5) v0
±
2 1
( −(3 / 5) v0
) −4(
g )( −h)
2
1
2( g )
2
2
5
0
9 25 gh 16 gh
( 3 / 5) 25gh
/ 8±
• +
25 8 8
t = . Substituting v
0
= 25gh
/ 8 gives t =
g
.
h
Collecting terms gives t: ( ) ( )
1 9 h
±
25h
=
1 3 h ± 5
meaningful and so
t 4
2h
g
t
2 2g
2g
2 2g
2g
= . Only the positive root is
= 4 .
(
0
cos ) ,
8 5
4
2g
=
25gh
4 h
= . Then, using x v α t x = ( )( ) h
3.49: The range for a projectile that lands at the same height from which it was launched
2
v sin 2α
g
0
is R = . Assuming α = 45°
, and R = 50 m, v0 = gR = 22 m/s .