Answer:
11.4 m/s
Step-by-step explanation:
Given that Jack drops a stone from rest off of the top of a bridge that is 21.2 m above the ground. After the stone falls 6.6 m, Jill throws a second stone straight down.
Since Both rocks hit the water at the exact same time, The final velocity V of Jack stone will be:
V^2 = U^2 + 2gH
Since the stone was dropped from rest, U = 0
V^2 = 2 × 9.8 × 21.2
V^2 = 415.52
V = 20.4 m/s
the initial velocity of Jill's rock will be found by using the same formula. Assuming that they both have the same final velocity V
V^2 = U^2 + 2gH
20.38^2 = U^2 + 2 × 9.8 × ( 21.2 - 6.6 )
415.52 = U^2 + 19.6 × 14.6
415.52 = U^2 + 286.16
U^2 = 415.52 - 286.16
U^2 = 129.36
U = 11.3736 m/s
Therefore, the initial velocity of Jill's rock is 11.4 m/s approximately