Here Eddy2,
This is how IFCR is going to get out of the mess that they're in, have at it.
Let ? be the angle at which the ball is thrown. Then the coordinates are given by
x = (v cos ?)t and y = (v sin ?)t - gt2/2. The ball reaches its maximum height at
t = v sin ?/g, so the length of the trajectory is
L = 2 Z v sin ?/g
0
sµ
dx
dt ¶2
+
µ
dx
dt ¶2
dt
= 2 Z v sin ?/g
0
q
(v cos ?)
2 + (v sin ? - gt)
2 dt
= 2v cos ?
Z v sin ?/g
0
s
1 + µ
tan ? -
gt
v cos ?
¶2
dt. (1)
Letting z = tan ? - gt/v cos ?, we obtain
L = -
2v
2
cos2
?
g
Z 0
tan ?
p
1 + z
2 dz. (2)
Letting z = tan a, and switching the order of integration, gives
L =
2v
2
cos2
?
g
Z ?
0
da
cos3 a
. (3)
You can either look up this integral, or you can derive it (see the remark at the end
of the solution). The result is
L =
2v
2
cos2
?
g
·
1
2
µ
sin ?
cos2 ?
+ ln µ
1 + sin ?
cos ?
¶¶
=
v
2
g
µ
sin ? + cos2
? ln µ
sin ? + 1
cos ?
¶¶ . (4)
As a double-check, you can verify that L = 0 when ? = 0, and L = v
2/g when
? = 90?
. Taking the derivative of eq. (4) to find the maximum, we obtain
0 = cos ? - 2 cos ? sin ? ln µ
1 + sin ?
cos ?
¶
+ cos2
?
µ
cos ?
1 + sin ?
¶
cos2
? + (1 + sin ?) sin ?
cos2 ?
.
(5)
This reduces to
1 = sin ? ln µ
1 + sin ?
cos ?
¶
.
Finally, you can show numerically that the solution for ? is ?0 ˜ 56.5
?
.
A few possible trajectories are shown below. Since it is well known that ? = 45?
provides the maximum horizontal distance, it follows from the figure that the ?0
yielding the arc of maximum length must satisfy ?0 = 45?
. The exact angle, however,
requires the above detailed calculation.
****This is how IFCR will resolve all their problems******
T.
