OK, here's what I did - refer to the attached drawing:
First, recognize that there are three phases to this problem. Phase 1: you have a mass on a string of length L which swings down in an arc of radius L. Phase 2: then the string hits the peg and you have the mass now moving in an arc of radius L-R. Phase 3: as the mass rises at some point the tension in the string goes to zero, and the mass is essentially lobbed back at the pin. For this phase it moves like a projectile under gravity.
1. In order for the mass to be lobbed back toward the pin the tension in the string must be zero, so that the mass moves in an arc under gravity. The tension in the string is:
} - mg \cdot sin \beta<br />
)
T is zero at the pont that the mass is no longer constraijed by the string. This gives you an expression for
in terms of
:
}<br />
)
2. Given an initial launch velocity of
, the time for the mass to move horizontally back to the peg is:
3. In order to precisely hit the peg, the mass has to fall vertically a distance
 sin \beta)
in that time T:
sin \beta = V_0 cos\beta T - \frac 1 2 g T^2<br />
)
4. Combine the equations from 2 and 3 to get
in terms of
:
} 2 \frac {cos^2 \beta} {sin \beta}<br />
)
5. Put the equation from step 1 into 4. This gets you a value for
purely in terms of g, L and R:
} {\sqrt 3}<br />
)
6. The height of the launch point is:
 sin \beta<br />
)
from step 1 and 5 we know that:
 sin \beta = \frac {V_0 ^2} g = \frac {(L-R)} {\sqrt 3}<br />
)
7. You know that when the mass is first released it must reach velocity
at this height as it swings downward. So use energy principles to calculate the height from which it must be released for this to occur:
Sub in
H from step 6, and
from step 5. Plug and chug, and you get:
<br />
)
