A 50 ball is fired horizontally with initial speed toward a 110 ball that is hanging motionless from a 1.0 -long string. The balls undergo a head-on, perfectly elastic collision, after which the 110 ball swings out to a maximum angle = 52.
What was V0?

h = L (1 - cos 52) = 1 * .384 = .384 m height to which m2 rises
1/2 m2 V2^2 = m2 g h
V2 = (2 * 9.8 * .384)^1/2 = 2.74 m/s
For an elastic collision the relative speed of separation equals
the relative speed of approach
V0 = V2 - V1 or you can go thru the math to show this
so V0 + V1 = V2 (I)
m1 V0 = m1 V1 + m2 V2 conservation of momentum
m1 (V0 - V1) = m2 V2 (II)
2 V0 = m2 / m1 * V2 + V2 = V2 (1 + m2 / m1) adding (I) and (II)
V0 = 1/2 * 2.74 * (1 + 110 / 50) = 4.38 m/sec