To solve this problem, we make use of the equations for linear motion. The relevant formula to use here is:
y = vi t + 0.5 a t^2Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â ---> 1
where,
 t = time, y = distance, a = acceleration = gravity, vi = initial velocity
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Let us say that the light rock is 1, and the heavy rock is 2. We know that the distance of the two rocks must be equal, therefore:
y1 = y2
vi1 t1 + 0.5 g t1^2 = vi2 t2 + 0.5 g t2^2
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From the given values, we know that rock 1 is simply dropped therefore vi1 = 0, therefore:
0.5 g t1^2 = vi2 t2 + 0.5 g t2^2Â Â Â Â Â Â Â Â Â Â Â Â Â ---> 2
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We also know that t1 = t2 + 0.5 or t2 = t1 – 0.5. Therefore we need first to find for the value of t1. By using equation 1:
55 = 0.5 (9.8) t1^2
t1^2 = 11.22
t1 = 3.35 s
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Therefore:
t2 = t1 – 0.5 = 3.35 – 0.5
t2 = 2.85 s
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 Going back to equation 2:
0.5 (9.8) (3.35)^2 = vi2 (2.85) + 0.5 (9.8) (2.85)^2
2.85 vi2 = 15.19
vi2 = 5.33 m/s
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Therefore he must throw the rock at an initial velocity of 5.33 m/s to reach the 1st rock.