Jack drops a stone from rest off of the top of a bridge that is 24.4 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? Assume upward is the positive direction and downward is negative. (Indicate the direction with the sign of your answer.)

dropped stone hits water:

24.4-4.9t^2=0
t=2.23

stone has fallen 6.6m:
4.9t^2 = 6.6
t = 1.16

So, Jill's thrown stone only has 2.23-1.16=1.07 seconds to hit the water:

24+1.07v-4.9*1.07^2 = 0
v = -17.19 m/s

Well, it seems like Jack and Jill are having quite the eventful day! Now, let's get to the question at hand.

We can start by figuring out the time it takes for Jack's stone to hit the water. We can use the equations of motion to solve for time. Using the equation:

h = ut + (1/2)gt^2

Where:
h = height (24.4 m)
u = initial velocity (0 m/s, as the stone is dropped from rest)
g = acceleration due to gravity (-9.8 m/s^2, since downward is negative)
t = time

Plugging the values into the equation, we get:

24.4 = 0 * t + (1/2) * (-9.8) * t^2

Simplifying, we have:

24.4 = -4.9t^2

Dividing both sides by -4.9, we find:

t^2 = -24.4 / -4.9
t^2 = 5

Taking the square root of both sides, we get:

t = √5

So, it takes √5 seconds for Jack's stone to hit the water.

Now, let's focus on Jill's rock. We know that it falls 6.6 m, and we want to find its initial velocity. Using the same equation of motion:

h = ut + (1/2)gt^2

Where:
h = height (6.6 m)
u = initial velocity (what we want to find)
g = acceleration due to gravity (-9.8 m/s^2, since downward is negative)
t = time (√5 seconds)

Plugging in the values, we have:

6.6 = u * √5 + (1/2) * (-9.8) * (√5)^2

Simplifying further, we get:

6.6 = u * √5 - 24.5

Rearranging the equation, we have:

u * √5 = 6.6 + 24.5
u * √5 = 31.1

Now dividing both sides by √5, we find:

u = 31.1 / √5

Calculating this, we get:

u ≈ 13.9 m/s

So, the initial velocity of Jill's rock is approximately 13.9 m/s, in the downward direction. Watch out for falling objects, folks!

To find the initial velocity of Jill's rock, we can use the equation of motion for both rocks.

For Jack's rock (dropped from rest):

Final velocity, Vf = ?
Initial velocity, Vi = 0 (since it is dropped from rest)
Acceleration, a = 9.8 m/s^2 (acceleration due to gravity)
Displacement, Δy = 24.4 m

The equation of motion is:

(Vf)^2 = (Vi)^2 + 2aΔy

Plugging in the values:

(Vf)^2 = (0)^2 + 2(9.8)(24.4)
Vf^2 = 2(9.8)(24.4)
Vf^2 = 476.96
Vf ≈ 21.84 m/s

Now, let's move on to Jill's rock:

Final velocity, Vf = 0 (since it hits the water)
Initial velocity, Vi = ?
Acceleration, a = 9.8 m/s^2 (acceleration due to gravity)
Displacement, Δy = 6.6 m

Using the equation of motion:

(Vf)^2 = (Vi)^2 + 2aΔy

Plugging in the given values:

(0)^2 = (Vi)^2 + 2(9.8)(6.6)
0 = (Vi)^2 + 128.76
(Vi)^2 = -128.76
Vi ≈ -11.35 m/s

Therefore, the initial velocity of Jill's rock is approximately -11.35 m/s. The negative sign indicates its downward direction.

To solve this problem, we can use the equations of motion for objects in free fall. Let's break down the problem into two parts: Jack's stone and Jill's stone.

1. Jack's stone:
- The stone is dropped from rest, which means its initial velocity (u) is 0 m/s.
- The bridge is 24.4 m above the ground, so the stone travels a total distance of 24.4 m.

We can use the following equation to calculate Jack's stone's time of flight (t) and the final velocity (v) at the end of its fall:

v^2 = u^2 + 2ad

Where:
- v is the final velocity (unknown).
- u is the initial velocity (0 m/s).
- a is the acceleration due to gravity (approximately 9.8 m/s^2).
- d is the distance traveled (24.4 m).

Plugging in the given values, the equation becomes:

v^2 = 0^2 + 2 * 9.8 * 24.4

Simplifying, we get:

v^2 = 2 * 9.8 * 24.4

Now, we can solve for v:

v = √(2 * 9.8 * 24.4)

Calculating the result:

v ≈ 21.949 m/s

2. Jill's stone:
- Jill throws her stone after Jack's stone has fallen by 6.6 m.
- We need to find the initial velocity (u) of Jill's stone.

We can use the following equation to calculate the initial velocity of Jill's stone:

d = ut + (1/2)at^2

Where:
- d is the distance traveled by Jill's stone (6.6 m).
- u is the initial velocity (unknown).
- a is the acceleration due to gravity (-9.8 m/s^2, because it is in the downward direction).
- t is the time for which the stone is in motion (unknown).

Since both stones hit the water at the same time, the time taken by Jill's stone will be the same as Jack's stone. So, we can use the calculated value of time from Jack's stone to solve for Jill's stone's initial velocity.

Rearranging the equation, we get:

u = (d - (1/2)at^2) / t

Plugging in the given values, the equation becomes:

u = (6.6 - (1/2) * 9.8 * t^2) / t

Now, let's substitute t with the value we found earlier for Jack's stone, which is approximately 21.949 m/s:

u = (6.6 - (1/2) * 9.8 * (21.949)^2) / 21.949

Calculating the result:

u ≈ -7.43 m/s (negative sign indicates the downward direction)

Therefore, the initial velocity of Jill's stone is approximately -7.43 m/s, indicating that it was thrown downward.

Jack: d=1/2 g t^2

not time when his stone is 6.6, is
t1=sqrt (6.6/4.9) compute that
Jill d=v*(t-t1) +1/2 g (t-t1)^2
Now you know d=24.4
solve for Jack time t
then, go to Jill equation, solve for v.