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. Both rocks hit the water at the exact same time. What was the initial velocity of Jill's rock

11.4 m/s

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

do you have a formula possibly