draw the figure:
initial PEnergy=mgl(1-cosTheta)
final KE=mgl(1-cosTheta)=1/2 mv^2
solve for v.
Then, that is also the hit ball velocity (conservation of momentum).
time to fall h,
h=1/2 g t^2
t=sqrt(2h/g)
distance=velocity*time
A small metal ball with a mass of m = 78.9 g is attached to a string of length l = 1.76 m. It is held at an angle of = 44.5° with respect to the vertical.
The ball is then released. When the rope is vertical, the ball collides head-on and perfectly elastically with an identical ball originally at rest. This second ball flies off with a horizontal initial velocity from a height of h = 2.72 m, and then later it hits the ground. At what distance x will the ball land?
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